8/12/2019 Fibonacci Year by Raah Sirus

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Uploaded first on - solar warrior : 1 muluc : day 66 of the organic spring equinox year - © 2012 Raah Sirus

 

Fibonacci year

:-) It is an absolute pleasure to be sharing this piece of the puzzle, as through its remembranceit will surely revolutionize our whole understanding of a great many things. What you are lookingat is a year based on the ratio's of the fibonacci spiral, which goes as follows:

 

0,1,1,2,3,5,8,13,21,34,55,89,144,...

 

The way that I have arranged the numbers is so that they create both an inner time matrixand outer time matrix.

 

Inner time (geometric) matrix: The first 6 numbers from 0-5, equal:

 zero = center from which everything is born, and to which everything returns

1+1+2+3+5= 12 = 12 month time systems

if the zero point (center) is included,

that would also allow for 13 month time systems.

 

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Outer time (organic) matrix: The next 7 numbers from 8-144, equal:

8+13+21+34+55+89+144 = 364 (+1) to complete the year = 365 days

 

In our current era, and in era's of the past, the organically measured seasons are made up of89 days and 93 days lengths. Anchoring the start of the outer time matrix at the point at whichthe numbers lock into a symmetry of 89 and 93, can be shown thus:

144 + 8 + 13 + 21 = 186 = ( 2 x 93 )

34 + 55 + 89 = 178 = ( 2 x 89 )

*it was certainly a sign post that one of the numbers already matched up i.e 89

see also this page: 8-fold-year 

 

Therefore plotting the numbers so that they sequence in the same way as our seasons currently

do, with spring = 93, summer = 93, autumn = 89(+1) and winter = 89, we can find avery interesting start point.

Axis of symmetry runs from summer solstice to winter solstice, the longer half (186 days)being spring and summer, and shorter half (178 days) autumn and winter.

Counting backwards from the autumn equinox, the point where long half and short half meet:

 

-21 -13 -8 = 42 days before autumn equinox does the outer time matrix lock in

autumn equinox 2012 = september 22nd @ 14:49 ( figures accoring to starcalc )

in the Mayan long count that is: 5 Oc ( 12.19.19.13.10 )

and 42 days previous to that is: 2 Lamat ( 12.19.19.11.8 )

in gregorian thats: august 11th 2012

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This is 132 days before the end of the current long count 20.12.2012 'the last day', and21.12.2012 'the first day' of next long count. Aug. 11th marks the exact point at which theSolar zenith occurs at Izapa in 2012, Izapa being the place where research says the longcount began.

see this page for more info:- solar-zenith-year

 

The chart gives the Zenith dates for Izapa, Palenque & Teotihuacan (years 100AD-2200AD)

 

It was at this poing in my own research that i've had to ask myself how much did the Mayansknow? the codes of the Fibonacci spiral that exist throughout the natural world and in the sizeand ratio of the human body... Leonardo Fibonacci lived between (1170-1250ad)...

 

And now it also describes the 'shape of the year' and organic year time matrix !

 

see this page for more info: organic-year

 

what a divine world we truly live in !

 

In lak'ech Raah

 

p.s. here is a graphic I created to summarize this particular aspect of the genius of the MayanTime Science, it also shows the relation of the Fibonacci ratio to 1 year, anchored at the

geometric solar zenith (did the Mayans also know of this ratio long before us?)

 

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8 + 13 + 21 + 34 + 55 = 131 (+1) = 132 days

number of days from geometric zenith to organic solstice 2012

 

Mayan Long Count lasts

5,125 years + 132 days

 

1,872,000 days

- 132 = 1871868 days

= 5125 x 365.2425 

 

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