C. America

© Copyr. 2008 by Dana Jay Smith  All Rights Reserved

Ancient Central American Riddles

Prerequisites:
  1. A Strange 12 Days of Christmas
  2. Enoch’s Riddles

In ancient central America there were various cycles in the apparent religious thinking.  Among them, there were  cycles of 91, 260, 819, 151840, and 175760 days. This article is about the year cycles of four ages:

676 + 364 + 312 + 676 = 4 Ages in 2028 “years”,

and,  4008 + 4010 + 4081 + 5026 = 4 Ages in 17125 “years” .

The word “years” is in parenthesis because perhaps these cycles have little to do with actual years? Or perhaps they do relate to actual years while also relating to symbolism about something else?

Let’s begin by looking at the 2028 year cycle as some sort of riddle (the Maya loved to create various riddles) that suggests that we separate out the numbers 312 and 364. Here is one way that creates the 175760 day religious cycle:

2028 – 312 = 1716,     2028 – 364 = 1664,  then

(1716 + 1664)(1716 – 1664) =  3380 x 52 = 175760.

Perhaps the number 1716 is a clue to the 171.6 day period explained in the 12 Days of Christmas article notes? This seems to perhaps be the case if the 2028 year period is actually 2028 synodic Moon periods:

2028 Synodic Moons = 2028 x 29.53059 = 59888.04 days,

349 periods of 171.6 days = 59888.40 days,

2192 Sidereal Moons = 2192 x 27.32166 = 59889.08 days.

Here we have a Moon cycle; however, large numbers of cycles are not all that meaningful because it is so easy to fit short cycles into large numbers. Therefore the 349 periods of 171.6 days sets up a more unlikely by chance Moon-Venus-Earth cycle of around 59888.5 days. It also fits into the concept of designing riddles to symbolize Venus-Moon influences in the mystical or religious sense. This alludes to the previously calculated correlation to the 175760 day religious cycle.

For the record, here is a  straight forward way of correlating the numbers 312 and 364 to the 260 x 676 = 175760 day  religious cycle:

10(Average[364, 312])(364 – 312) = 260×676 = 175760.

Due to multiplying by 10, the resulting 3380×52 numbers end up being identical to the previous calculation.

I have noticed something else that relates to the 2028, 2192 numbers along with the number 4:

(2028 + 2192)(2192 – 2028)/4 = 173020.

173020 days is 770 Venus years, 712 Venus rotations, and 5859 synodic Moon periods. This is a very accurate cycle, but I am not sure what to think about it? Once again, this does relate to Venus-Moon symbols; but not to any other central American symbols (except 2028). It is interesting to notice that the sum and difference of numbers relating to a Moon-Venus-Earth cycle conspire to produce a Venus-Moon cycle? Perhaps the following is a clue to yet another riddle:

175760  –  173020 = 2740,

where 2740 is the 3 digit reciprocal of 365 or 365.01. This relates to the sum of the other 4 Age riddle:

27400 / 1.6 = 17125.

Let’s look at what can be done with the 17125 year central American cycle.  Recall:

4008 + 4010 + 4081 + 5026 = 17125 “years”.

The sum is very suspicious because the reciprocal of the VES (Venus-Earth synodic period) is about 17125.63. The number 17125 is the truncated approximation. It does have the nice property of creating various double four digit sequences for symbolic purposes. Here is an example:

394784 x 17125 = 6760676000,

which is a crude approximation to Enoch’s Ratio (see Enoch’s Riddles). The number 394784 = 584 x 676, where the Mayan Agricultural Cycle was 584 times 260 days, and of course, 676 x 260 days was the religious cycle.

Anyway, the fact that 17125 is close to the reciprocal of the VES makes this “smell” like a riddle. Here is what can be done with these numbers:

(5026² – 4008²)(5026² + 4008²)(4010 x 4081) / (17125.136 Million)

= CP =  603550  x 601730.

Notice that we only had to adjust the number 17125 up very slightly to bring the result into the exact Book of Numbers 12 tribe census product.

Eliminating the slight adjustment of the 17125 sum might be the most interesting part of this riddle.

.

While this solution may look like a numerical coincidence, such is very very unlikely because each section of the calculation relates to Enoch’s Riddles and the Special Venus Triangle or SVT (see the notes in The Strange 12 Days of Christmas article):

(5026² – 4008²) / 171.25136 = 13×17 x 242.9973634.

Notice the 13×17 Venus rotations or approximately 13 x 17 x Venus Year divided by SinØ ( this should be familiar to you if you have done your homework; this is a SVT calculation):

(5026² – 4008²) / 171.25136 = 13×17 x 224.7006/SinØ.

The central numbers 4010 and 4081 are multiplied together:

4010 x 4081 = 16364810.

Recall this number to be near one of the numbers that converts between Enoch’s Ratio and Enoch’s Sum:

Enoch’s Sum x Enoch’s Ratio = 163652.41.

Notice the analogy, both the 4010, 4081 and Ens, Enr calculations are products.  4010 x 4081  =  Enr x 224.7 x 295.53819 x 36449.78 x 100, which is close to 100 x Literal Enoch’s Sum x Enoch’s Ratio. It is possible that the 4010 x 4081 number is the correct number for 10 literal Enr times 10 literal Ens. Such is the case for a Venus rotation of 243.003 days, which is very close to the nominal value of 243.

Next is the approximate reciprocal of Enoch’s Sum:

5026² + 4008²  =  41324740.

Let’s look at the product:

(5026² + 4008²)(4010 x 4081) / 10¹³  =  67.62715184 degrees.

Here we have that necessary angle near 67.63 degrees. All of these various approximations relating to Enoch’s Riddles and the Special Venus Triangle (SVT) have conspired together to produce the 12 tribe census product. Because the numbers are on the high side, we end up with a number close to 67.63, rather than the Enoch Ratio symbol:

10^7 x 67.62715184(5026² – 4008²)/17125.136 = CP.

The fact that the various parts of the riddle correlate properly makes this very unlikely to have happened by pure coincidence. Let’s look at what is required to eliminate the adjustment of the 17125 sum. The numerator of the calculation factors into 3 parts, A, B, and C:

(A)(B)(C)/(17125 Million) = CP = 603550 x 601730.

We can subtract a correction term, denoted by ε, from one of the factors, and then solve the resulting equation for the amount of correction needed within that particular factor to generate the exact CP. For example:

(A – ε)(B)(C)/17125 Million = CP, solve for ε.

Hopefully, if the riddle is properly designed, the required correction will be close to one of the small numbers that can be generated from the difference of various numbers available to us (only a few such numbers are available). Here are the results:

  1. (A – ε) factor = ( 5026² + 4008²  – ε )   –>  ε = 329.49
  2. (B – ε) factor = ( 4010 x 4081 – ε )  —–>  ε = 130.48
  3. (C – ε) factor = ( 5026² – 4008² – ε)  —–>  ε = 73.33

The riddle has apparently been designed to be corrected by the number (4081 – 4008) = 73, where 73 x 5 = 365, and 73 x 8 = 584, or 73/2 = 36.5 which equals (4081 – 4008)/(4010 – 4008) = 36.5.  This is a good choice relative to the 365 number in Enochian Symbolism. Here is the final result:

(5026²+4008²)(4010×4081)(5026² – 4008² – 4081 + 4008)/17125 Million

is equal to  (603550.0107)(601730.0107) = CP.  Due to only having a whole number available for the symbolic correction, there is a very small error involved.  We now have the riddle solved while only using the numbers 4008, 4010, 4081, 5026, and the sum of these numbers.

NOTES:

1) It happens that we can use the symbolic numbers 312 and 676 to bring an Enoch Sum approximation into exactness:

Ens = 242060 = 10¹³/(5026² + 4010² – 6x7x676 – 312).

This result is exact in the practical sense;  312.6 brings 10 decimal place accuracy.

2) Here is a bit of “7 come 11” luck:

4 x 4008x4010x4081x5026 / (7 x 11 x 10^9) = 17125.

The exact result is 17124.9941.

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