### Corrections

o **Corrections For Decimal Date Conversion Tables**

While every effort has been made to assure the accuracy of the algorithms and calculations for the time Code Calculator, an error has been discovered in the decimal-to-month conversion tables. While the error has minimal impact upon the calculations, it does change the month in some cases. And, with that change, some of the Time Code calculations are now even closer than before to the actual dates when their cycles repeated. Please refer to the tables below and use these decimal equivalents in place of those in the first edition of Fractal Time.

** **

* Note 2***:** Conversion of decimals to months. Some of the calculated dates create numbers to the right of the decimal point. These are portions (fractions) of the year indicated and may be converted into the corresponding month for greater accuracy using the following formula.

· (Number to right of decimal ÷ 12) x 100 = percent of the year

**Example:** Year 2001.8 translates to (8 ÷ 12) x 100 = 66.66 percent of the year or 2001.67 (August of 2001). The key here is to think of the decimal that is expressed in tenths, as a portion of the 12 months that are possible. For convenience, I have included a reference chart of the decimal-to-month calculations for numbers 1 to 12 below:

__Portion of Year__ __Equivalent Month__

0.08 January

0.16 February

0.25 March

0.33 April

0.42 May

0.50 June

0.58 July

0.67 August

0.75 September

0.83 October

0.91 November

0.99 December

*Note 3:** *These formulas calculate *zones of time* that make the events possible, rather than the specific date and time that an event will take place. So in our example of September 11, 2001, as a repeat of the cycle that began in 1941, the calculations show a 30-day lag between the Time Code Calculation and the actual event. The key is that the Calculator clearly pinpoints the time(s) of the repeating pattern within the greater 5,125-year cycle.

As the discussion of choice points in chapter 7 illustrates, human choices can alter the course of events, even when the conditions that support those events are present. And this fact is precisely why the Time Code Calculations so valuable. It gives us the heads up as to when we can expect such conditions to be present.

o **Corrections For Examples in Appendices**

**Mode 1: When Can We Expect Something that Has Happened in the Past to Happen Again?
**

To answer this question we need two pieces of information:

· *Input 1*: The target date in the past when an obvious pattern (the seed) occurred

· *Input 2*: The total length of the cycle that tells us where we are in present time

**We’ll be using four examples to demonstrate time code calculations for repeated conditions.**

**Example 1—Find: When can we expect the first cyclic conditions of “surprise” and “attack” on America to repeat?
**·

*Input 1*:

**The target year and month in the past when the first obvious pattern of “surprise” and “attack” upon America (the seed) occurred**

**: 1941.12 (December 1941)**

·

*Input 2*:

**The total length of the cycle that tells us where we are in present time:**

**5,125 years**

Step 1: Identify modern (Gregorian) date of seed event (S_{1}).

** 1941.99
**Step 2: Convert event (S

_{1}) to absolute date (A

_{1}).

**1941.99 + 3113 = 5054.99**(A

_{1})

**Step 3: Calculate lapsed portion of cycle (L**

_{1})

**as a ratio of (A**

_{1})

**/**total cycle length 5125.

**5054.99/5125 = .986**(L

_{1})

**Step 4: Calculate phi (L**

_{1phi}) of lapsed cycle (L

_{1}).

**(.618) x .986 = .609**(L

_{1phi})

Step 5: Calculate cycle balance (B

_{1}) as total cycle length (A

_{1}).

**5125 – 5054.99 = 70.01 years**(B

_{1})

**Step 6: Apply phi ratio of lapsed cycle (L**

_{1}) to the balance of the cycle (B

_{1}) to find the interval in years between the seed date and the next time it repeats (I

_{1}).

**(70.01) x (.609) = 42.64 years**(I

_{1})

Step 7: Add the interval (I

_{1}) to the original seed date to find the next repeat (new seed date).

**5054.99 + 42.64 = 5097.63**

Step 8: Convert back to Gregorian date.

**5097.63 – 3113 = 1984.63 (August 1984)**

*Meaning*: This date translates to August 1984. The date for the destruction of KAL flight 007 and the events described in Chapter 1 occurred in September 1983, only eleven months earlier. The time span

*between*September 1983 and February 1984 is documented as one of the tensest periods of the war between the two superpowers. Post-Cold War documents reveal that it was precisely during this time frame, and within 6 months of the time predicted by the Time Code Calculator, that a pre-emptive nuclear strike was being planned against the United States.

The time code calculations demonstrate that the plans for a surprise attack on America—the

*first fractal*of the pattern that was created in 1941—are part of the cyclic pattern that can be known and calculated. As described in the text and shown in the following example, the second fractal pattern occurred in September 2001.

*Example 2—Find: When can we expect the ***second cyclic conditions of “surprise” and “attack” on America to repeat?**

*·*

*Input 1*:

**The first target year following the seed of 1941 when an obvious pattern of “surprise” and “attack” upon America (the seed) occurred**

**: 1984.63 (August 1984)**

·

*Input 2*:

**The total length of the cycle that tells us where we are in present time:**

**5,125 years**

Step 1: Identify modern (Gregorian) date of seed event (S_{1}).

** 1984.63
**Step 2: Convert event (S

_{1}) to absolute date (A

_{1}).

**1984.63 + 3113 = 5097.63**(A

_{1})

**Step 3: Calculate lapsed portion of cycle (L**

_{1})

**as a ratio of (A**

_{1})

**/**total cycle length 5125.

**5097.63/5125 = .995**(L

_{1})

**Step 4: Calculate phi (L**

_{1phi}) of lapsed cycle (L

_{1}).

**(.618) X .995 = .615**(L

_{1phi})

Step 5: Calculate cycle balance (B

_{1}) as total cycle length (A

_{1}).

**5125 – 5097.63 = 27.37 years**(B

_{1})

**Step 6: Apply phi ratio of lapsed cycle (L**

_{1}) to the balance of the cycle (B

_{1}) to find the interval between the seed date and the next time it repeats (I

_{1}).

**(27.37) x (.615) = 16.83 years**(I

_{1})

Step 7: Add the interval (I

_{1}) to the original seed date (S

_{1}) to find the next repeat (new seed date).

**5097.63 + 16.83= 5114.46**

Step 8: Convert back to Gregorian date.

**5114.46 – 3113 = 2001.46 (June 2001)**

** ***Meaning*: This date translates to June 2001. It falls well within the range of the time that the attack is believed to have been in the planning phase, and it is within three months of the date that the World Trade Center and Pentagon attacks actually occurred. There is only a 1 in 61,500, or .0000162 percent, chance of determining that 2001 would be the year within the present world age cycle of such an attack.

** **

**Example 3—Find: When can we expect the third cyclic conditions of “surprise” and “attack” on America to repeat?
**·

*Input 1:*

**The first target year following the seed of 1941 when an obvious pattern of “surprise” and “attack” upon America (the seed) occurred**

**: 2001.46 (June 2001)**

·

*Input 2:*

**The total length of the cycle that tells us where we are in present time:**

**5,125 years**

Step 1: Identify modern (Gregorian) date of seed event (S_{1}).

** 2001.46
**Step 2: Convert event (S

_{1}) to absolute date (A

_{1}).

**2001.46 + 3113 = 5114.46**(A

_{1})

**Step 3: Calculate lapsed portion of cycle (L**

_{1})

**as a ratio of (A**

_{1})

**/**total cycle length 5125.

**5114.46/5125 = .998**(L

_{1})

**Step 4: Calculate phi (L**

_{1phi}) of lapsed cycle (L

_{1}).

**(.618) x .998 = .617**(L

_{1phi})

Step 5: Calculate cycle balance (B

_{1}) as total cycle length (A

_{1}).

**5125 – 5114.46 = 10.54 years**(B

_{1})

**Step 6: Apply phi ratio of lapsed cycle (L**

_{1}) to the balance of the cycle (B

_{1}) to find the interval between the seed date and the next time it repeats (I

_{1}).

**(10.54) x (.617) = 6.50 years**(I

_{1})

Step 7: Add the interval (I

_{1}) to the original seed date (S

_{1}) to find the next repeat (new seed date).

**5114.46 + 6.50 = 5120.96**

Step 8: Convert back to Gregorian date.

**5120.91 – 3113 = 2007.96 (December 2007)**

**Example 4—Find: When can we expect the fourth cyclic conditions of “surprise” and “attack” on America to repeat?**

·

*Input 1*:

**The first target year following the seed of 1941 when an obvious pattern of “surprise” and “attack” upon America (the seed) occurred**

**: 2007.96 (December 2007)**

·

*Input 2*:

**The total length of the cycle that tells us where we are in present time:**

**5,125 years**

Step 1: Identify modern (Gregorian) date of seed event (S

_{1}).

**2007.96**

Step 2: Convert event (S

_{1}) to absolute date (A

_{1}).

**2007.96 + 3113 = 5120.96**(A

_{1})

**Step 3: Calculate lapsed portion of cycle (L**

_{1})

**as a ratio of (A**

_{1})

**/**total cycle length 5125.

**5120.96/5125 = .999**(L

_{1})

**Step 4: Calculate phi (L**

_{1phi}) of lapsed cycle (L

_{1}).

**(.618) x .999 = .617**(L

_{1phi})

Step 5: Calculate cycle balance (B

_{1}) as total cycle length (A

_{1}).

**5125 – 5120.96 = 4.04 years**(B

_{1})

**Step 6: Apply phi ratio of lapsed cycle (L**

_{1}) to the balance of the cycle (B

_{1}) to find the interval between the seed date and the next time it repeats (I

_{1}).

**(4.04) x (.617) = 2.49 years**(I

_{1})

Step 7: Add the interval (I

_{1}) to the original seed date (S

_{1}) to find the next repeat (new seed date).

**5120.96 + 2.49 = 5123.45**

Step 8: Convert back to Gregorian date.

** 5123.45 – 3113 = 2010.45 (June 2010)**

** ***Meaning*: This date translates to June 2010. It is the return of the seed cycle planted in 1941, so it is also identifies the greatest opportunity to heal the conditions that led to the events of the original seed. The weeks and months preceding this date hold the greatest opportunity for the conscious easing of tension and creation of peace from the last repeat of 2007, until the next return.

** **

** **

**Mode 2: What Date in the Past Holds the Conditions We Can Expect for a Date in the Future?
**

To answer this question we need two pieces of information:

· *Input 1*: The target date in the future that is in question.

· *Input 2*: The total length of the cycle that tells us where we are in present time.

__ __

*The Time Code Algorithm Described: ***Always apply these same five steps.**

*The Time Code Algorithm Described:*

Step 1: Identify modern (Gregorian) date of target event.

Step 2: Convert the Gregorian date to an “absolute” date in terms of the total cycle for ease of calculations (optional).

Step 3: Calculate the phi ratio of the absolute date (multiply by .618)

Step 4: Subtract the phi ratio date from the target date.

Step 5: Convert back to Gregorian date.

*The Time Code Algorithm Applied to the 2012 End Date: ***Here’s an example of how to search backwards through time for seed conditions.**

*The Time Code Algorithm Applied to the 2012 End Date:*

**Example 1—Find: What date in the past holds the conditions can we expect for the 2012 end date?
**·

*Input 1*:

**The target date in question:**

**2012**·

*Input 2*:

**The total length of the cycle that tells us where we are in present time:**

**5,125 years**

Step 1: Identify modern (Gregorian) date of target date (T_{1}).

** 2012
**Step 2: Convert target date (T

_{1}) to absolute date (A

_{1}).

**2012 + 3113 = 5125**(A

_{1})

**Step 3: Calculate phi (L**

_{1phi}) of Absolute date (A

_{1}).

**(.618) x 5125 = 3167.25**(L

_{1phi})

Step 4: Subtract the phi ratio date (L

_{1phi}) from the target date (A

_{1}).

**5125 – 3167.25 = 1957.75**

Step 5: Convert back to Gregorian date.

**1957.75 – 3113 = -1155.25 (March –1155 b.c.)**

** ***Meaning*: The result of this calculation is a negative number, indicating that it is a date before the time of Christ (b.c.) in the historic notation. As noted in Chapter 6, this is precisely the year that witnessed the collapse of one of the greatest civilizations of the past, Egypt’s 20th Dynasty. The parallels between the conditions of 1155 b.c. and the 2012 close of the present great cycle are unmistakable. By applying the nature’s language of cycles, the Time Code Calculator identifies the single date in the past 5,125 years that holds the key to what we can expect for our near future.

Appendix B

Global Flashpoints for the Future

We have seen how the cycles of time and events repeat in rhythmic intervals that follow the mysterious number phi (.618). Knowing this, we can find with the seed events that posed the greatest threats to our world in the 20th century to discover when we can expect the conditions they created to appear again. We can use our knowledge of such conditions as a window of opportunity to avoid in the present the outcomes of the past.

Knowing when the patterns of the last century’s global wars are primed to repeat gives us the edge to avoid new conflicts based upon the old patterns. If we know that we’re in a year when the cycle that led to World War II is repeating, for example, then we also know that we’re in the time when it is wise to use a little extra care and sensitivity when dealing with the inevitable disagreements that arise between nations over resources, borders and human rights.

Using Mode 1 of the Time Code Calculator we can identify when such a cycle begins in order to calculate when the conditions it created will repeat in our future. What’s important here is that the seed event is what begins the cycle.

** **

Mode 1: When Can We Expect Something that Has Happened in the Past to Happen Again?

To answer this question we need two pieces of information:

· *Input 1*: The target date in the past when an obvious pattern (the seed) occurred

· *Input 2*: The total length of the cycle that tells us where we are in present time

We’ll be using nine examples to demonstrate time code calculations for future events.

**Example 1—Find: First cycle repeat date for the conditions of 1945 (atomic weapons and end of war)
**·

*Input 1*:

**The seed date for global war in the 20th century:**

**1945**

·

*Input 2*:

**The total length of the cycle that tells us where we are in present time:**

**5,125 years**

Step 1: Identify modern (Gregorian) date of seed event (S_{1}).

** 1945
**Step 2: Convert event (S

_{1}) to absolute date (A

_{1}).

**1945 + 3113 = 5058**(A

_{1})

**Step 3: Calculate lapsed portion of cycle (L**

_{1})

**as a ratio of (A**

_{1})

**/**total cycle length 5125.

**5058/5125 = .987**(L

_{1})

**Step 4: Calculate phi (L**

_{1phi}) of lapsed cycle (L

_{1}).

**(.618) x .987 = .610**(L

_{1phi})

Step 5: Calculate cycle balance (B

_{1}) as total cycle length (A

_{1}).

**5125 – 5058 = 67 years**(B

_{1})

**Step 6: Apply phi ratio of lapsed cycle (L**

_{1}) to the balance of the cycle (B

_{1}) to find the interval in years between the seed date and the next time it repeats (I

_{1}).

**(67) x (.610) = 40.87 years**(I

_{1})

Step 7: Add the interval (I

_{1}) to the original seed date to find the next repeat (new seed date).

**5058 + 40.87 = 5098.87**

Step 8: Convert back to Gregorian date.

**5098.87 – 3113 = 1985.87 (November 1985)**

**Example 2—Find: Second cycle repeat date for the conditions of 1945 (atomic weapons and end of war)
**·

*Input 1*:

**The seed date for global war in the 20th century:**

**1985.87**

·

*Input 2*:

**The total length of the cycle that tells us where we are in present time:**

**5,125 years**

Step 1: Identify modern (Gregorian) date of seed event (S_{1})

** 1985.87
**Step 2: Convert event (S

_{1}) to absolute date (A

_{1})

**1985.87 + 3113 = 5098.87**(A

_{1})

**Step 3: Calculate lapsed portion of cycle (L**

_{1})

**as a ratio of (A**

_{1})

**/**total cycle length 5125

**5098.87/5125 = .995**(L

_{1})

**Step 4: Calculate phi (L**

_{1phi}) of lapsed cycle (L

_{1})

**(.618) x .995 = .615**(L

_{1phi})

Step 5: Calculate cycle balance (B

_{1}) as total cycle length - (A

_{1})

**5125 – 5098.87 = 26.13 years**(B

_{1})

**Step 6: Apply phi ratio of lapsed cycle (L**

_{1}) to the balance of the cycle (B

_{1}) to find the interval in years between the seed date and the next time it repeats (I

_{1})

**(26.13) X (.615) = 16.07 years**(I

_{1})

Step 7: Add the interval (I

_{1}) to the original seed date to find the next repeat (new seed date)

**5098.87 + 16.07 = 5114.94**

Step 8: Convert back to Gregorian date

**5114.94 – 3113 = 2001.94 (December 2001)**

**Example 3—Find: Third cycle repeat date for the conditions of 1945 (atomic weapons and end of war)
**·

*Input 1*:

**The seed date for global war in the 20th century:**

**2001.94**

·

*Input 2*:

**The total length of the cycle that tells us where we are in present time:**

**5,125 years**

Step 1: Identify modern (Gregorian) date of seed event (S_{1})

** 2001.94
**Step 2: Convert event (S

_{1}) to absolute date (A

_{1})

**2001.94 + 3113 = 5114.94**(A

_{1})

**Step 3: Calculate lapsed portion of cycle (L**

_{1})

**as a ratio of (A**

_{1})

**/**total cycle length 5125

**5114.94/5125 = .998**(L

_{1})

**Step 4: Calculate phi (L**

_{1phi}) of lapsed cycle (L

_{1})

**(.618) x .998 = .617**(L

_{1phi})

Step 5: Calculate cycle balance (B

_{1}) as total cycle length - (A

_{1})

**5125 – 5114.94 = 10.06 years**(B

_{1})

**Step 6: Apply phi ratio of lapsed cycle (L**

_{1}) to the balance of the cycle (B

_{1}) to find the interval in years between the seed date and the next time it repeats (I

_{1})

**(10.06) x (.617) = 6.21 years**(I

_{1})

Step 7: Add the interval (I

_{1}) to the original seed date to find the next repeat (new seed date)

**5114.94 + 6.21 = 5121.15**

Step 8: Convert back to Gregorian date

**5121.15 – 3113 = 2008.15 (February 2008)**

** Example 4—Find: Fourth cycle repeat date for the conditions of 1945 (atomic weapons and end of war)**·

*Input 1*:

**The Seed Date For Global War In the 20th century:**

**2008.15**

·

*Input 2*:

**The total length of the cycle that tells us where we are in present time:**

**5,125 years**

Step 1: Identify modern (Gregorian) date of seed event (S_{1})

** 2008.15
**Step 2: Convert event (S

_{1}) to absolute date (A

_{1})

**2008.15 + 3113 = 5121.15**(A

_{1})

**Step 3: Calculate lapsed portion of cycle (L**

_{1})

**as a ratio of (A**

_{1})

**/**total cycle length 5125

**5121.15/5125 = .999**(L

_{1})

**Step 4: Calculate phi (L**

_{1phi}) of lapsed cycle (L

_{1})

**(.618) x .999 = .617**(L

_{1phi})

Step 5: Calculate cycle balance (B

_{1}) as total cycle length - (A

_{1})

**5125 – 5121.15 = 3.85 years**(B

_{1})

**Step 6: Apply phi ratio of lapsed cycle (L**

_{1}) to the balance of the cycle (B

_{1}) to find the interval in years between the seed date and the next time it repeats (I

_{1})

**(3.85) X (.617) = 2.38 years**(I

_{1})

Step 7: Add the interval (I

_{1}) to the original seed date to find the next repeat (new seed date)

**5121.15 + 2.38 = 5123.53**

Step 8: Convert back to Gregorian date

**5123.53 – 3113 = 2010.53 (July 2010)**

**Example 5—Find: First cycle repeat date for the conditions of global war that began in 1914
**·

*Input 1*:

**The Seed Date For Global War In the 20th century:**

**1914**

·

*Input 2*:

**The total length of the cycle that tells us where we are in present time:**

**5,125 years**

Step 1: Identify modern (Gregorian) date of seed event (S_{1})

** 1914
**Step 2: Convert event (S

_{1}) to absolute date (A

_{1})

**1914 + 3113 = 5027**(A

_{1})

Step 3: Calculate lapsed portion of cycle (L

_{1})

**as a ratio of (A**

_{1})

**/**total cycle length 5125

**5027/5125 = .981**(L

_{1})

Step 4: Calculate phi (L

_{1phi}) of lapsed cycle (L

_{1})

**(.618) x .981 = .606**(L

_{1phi})

Step 5: Calculate cycle balance (B

_{1}) as total cycle length - (A

_{1})

**5125 – 5027 = 98 years**(B

_{1})

Step 6: Apply phi ratio of lapsed cycle (L

_{1}) to the balance of the cycle (B

_{1}) to find the interval in years between the seed date and the next time it repeats (I

_{1})

**(98) X (.606) = 59.39 years**(I

_{1})

Step 7: Add the interval (I

_{1}) to the original seed date to find the next repeat (new seed date)

**5027 + 59.39 = 5086.39**

Step 8: Convert back to Gregorian date

**5086.39 – 3113 = 1973.39 (May 1973)**

**Example 6—Find: Second cycle repeat date for the conditions of global war that began in 1914
**·

*Input 1*:

**The Seed Date For Global War In the 20th century:**

**1973.39**

·

*Input 2*:

**The total length of the cycle that tells us where we are in present time:**

**5,125 years**

Step 1: Identify modern (Gregorian) date of seed event (S_{1})

** 1973.39
**Step 2: Convert event (S

_{1}) to absolute date (A

_{1})

**1973.39 + 3113 = 5086.39**(A

_{1})

**Step 3: Calculate lapsed portion of cycle (L**

_{1})

**as a ratio of (A**

_{1})

**/**total cycle length 5125

**5086.39/5125 = .992**(L

_{1})

**Step 4: Calculate phi (L**

_{1phi}) of lapsed cycle (L

_{1})

**(.618) x .992 = .613**(L

_{1phi})

Step 5: Calculate cycle balance (B

_{1}) as total cycle length - (A

_{1})

**5125 – 5086.39 = 38.61 years**(B

_{1})

**Step 6: Apply phi ratio of lapsed cycle (L**

_{1}) to the balance of the cycle (B

_{1}) to find the interval in years between the seed date and the next time it repeats (I

_{1})

**(38.61) x (.613) = 23.67 years**(I

_{1})

Step 7: Add the interval (I

_{1}) to the original seed date to find the next repeat (new seed date)

**5086.39+ 23.67 = 5110.06**

Step 8: Convert back to Gregorian date

**5110.06 – 3113 = 1997.06 (January 1997)**

**Example 7–Find: Third cycle repeat date for the conditions of global war that began in 1914
**·

*Input 1*:

**The seed date for global war in the 20th century:**

**1997.02**

·

*Input 2*:

**The total length of the cycle that tells us where we are in present time:**

**5,125 years**

Step 1: Identify modern (Gregorian) date of seed event (S_{1})

** 1997.06
**Step 2: Convert event (S

_{1}) to absolute date (A

_{1})

**1997.06 + 3113 = 5110.06**(A

_{1})

Step 3: Calculate lapsed portion of cycle (L

_{1})

**as a ratio of (A**

_{1})

**/**total cycle length 5125

**5110.06/5125 = .997**(L

_{1})

Step 4: Calculate phi (L

_{1phi}) of lapsed cycle (L

_{1})

**(.618) x .997 = .616**(L

_{1phi})

Step 5: Calculate cycle balance (B

_{1}) as total cycle length - (A

_{1})

**5125 – 5110.06 = 14.94 years**(B

_{1})

Step 6: Apply phi ratio of lapsed cycle (L

_{1}) to the balance of the cycle (B

_{1}) to find the interval in years between the seed date and the next time it repeats (I

_{1})

**(14.94) X (.616) = 9.20 years**(I

_{1})

Step 7: Add the interval (I

_{1}) to the original seed date to find the next repeat (new seed date)

**5110.06 + 9.20 = 5119.26**

Step 8: Convert back to Gregorian date

**5119.26 – 3113 = 2006.26 (April 2006)**

**Example 8–Find: Fourth cycle repeat date for the conditions of global war that began in 1914
**·

*Input 1*:

**The Seed Date For Global War In the 20th century:**

**2006.26**

·

*Input 2*:

**The total length of the cycle that tells us where we are in present time:**

**5,125 years**

Step 1: Identify modern (Gregorian) date of seed event (S_{1})

** 2006.26
**Step 2: Convert event (S

_{1}) to absolute date (A

_{1})

**2006.26 + 3113 = 5119.26**(A

_{1})

**Step 3: Calculate lapsed portion of cycle (L**

_{1})

**as a ratio of (A**

_{1})

**/**total cycle length 5125

**5119.26/5125 = .999**(L

_{1})

**Step 4: Calculate phi (L**

_{1phi}) of lapsed cycle (L

_{1})

**(.618) x .999 = .617**(L

_{1phi})

Step 5: Calculate cycle balance (B

_{1}) as total cycle length - (A

_{1})

**5125 – 5119.26 = 5.74 years**(B

_{1})

**Step 6: Apply phi ratio of lapsed cycle (L**

_{1}) to the balance of the cycle (B

_{1}) to find the interval in years between the seed date and the next time it repeats (I

_{1})

**(5.74) X (.617) = 3.54 years**(I

_{1})

Step 7: Add the interval (I

_{1}) to the original seed date to find the next repeat (new seed date)

**5119.26 + 3.54 = 5122.80**

Step 8: Convert back to Gregorian date

**5122.80 – 3113 = 2009.80 (October 2009)**

**Example 9—Find: Fifth cycle repeat date for the conditions of global war that began in 1914
**·

*Input 1*:

**The Seed Date For Global War In the 20th century:**

**2009.80**

·

*Input 2*:

**The total length of the cycle that tells us where we are in present time:**

**5,125 years**

Step 1: Identify modern (Gregorian) date of seed event (S_{1})

** 2009.80
**Step 2: Convert event (S

_{1}) to absolute date (A

_{1})

**2009.80 + 3113 = 5122.80**(A

_{1})

**Step 3: Calculate lapsed portion of cycle (L**

_{1})

**as a ratio of (A**

_{1})

**/**total cycle length 5125

**5122.80/5125 = 1.00**(L

_{1})

**Step 4: Calculate phi (L**

_{1phi}) of lapsed cycle (L

_{1})

**(.618) X 1.00 = .618**(L

_{1phi})

Step 5: Calculate cycle balance (B

_{1}) as total cycle length - (A

_{1})

**5125 – 5122.80 = 2.20 years**(B

_{1})

**Step 6: Apply phi ratio of lapsed cycle (L**

_{1}) to the balance of the cycle (B

_{1}) to find the interval in years between the seed date and the next time it repeats (I

_{1})

**(2.20) X (.618) = 1.36 years**(I

_{1})

Step 7: Add the interval (I

_{1}) to the original seed date to find the next repeat (new seed date)

**5122.80 + 1.36 = 5124.16**

Step 8: Convert back to Gregorian date

**5124.16 – 3113 = 2011.16 (February 2011)**

__ __

Appendix C

Reference Dates for 2012 Conditions

We will use Mode 2 of the Time Code Calculator to pinpoint the times in the past when the conditions of the 2012 end date last appeared. With these dates in mind, we can then use the template created in Chapter 6 to make a meaningful comparison of those times from two different cycles—the 5,125-year world age cycle and the 25,625 precessional cycle—to give us an idea of what we can expect for 2012. Following the examples of Appendices 1 and 2, the steps of the process are described in words, followed by the calculations themselves.

** **

**Mode 2:What Date in the Past Holds the Conditions We Can Expect for the Future?
**

** **To answer this question we need two pieces of information:

· *Input 1*: The target date in the future that is in question.

· *Input 2*: The total length of the cycle that tells us where we are in present time.

** Example 1–Find: What date in the 5,125-year world age cycle holds the conditions we can expect for the 2012 end date?**·

*Input 1:*

**The target date in question:**

**2012**·

*Input 2*:

**The total length of the cycle that tells us where we are in present time:**

**5,125 years**

Step 1: Identify modern (Gregorian) date of target date (T_{1})

** 2012
**Step 2: Identify total length of cycle in absolute years (C

_{1})

**5125**

Step 3: Calculate phi (L

_{1phi}) of total cycle (C

_{1})

**(.618) x 5125 = 3167.25**(L

_{1phi})

Step 4: Subtract the cycle’s phi point (L

_{1phi}) from target date (T

_{1})

**2012–3167.25 = –1155.25 b.c.**

*Example 2—Find: What date in the 25,625-year precessional cycle marks the conditions can we expect for the 2012 end date?*

· *Input 1*:** **The target date in question: **2012**·

*Input 2*:

**The total length of the cycle that tells us where we are in present time:**

**25,626 years**

Step 1: Identify modern (Gregorian) date of target date (T_{1})

** 2012
**Step 2: Identify total length of cycle in absolute years (C

_{1})

**25,625**

Step 3: Calculate phi (L

_{1phi}) of Total Cycle (C

_{1})

**(.618) x 25,625 = 15,836.25**(L

_{1phi})

Step 4: Subtract the cycle’s phi point (L

_{1phi}) from target date (T

_{1})

**2012 – 15836.25 = –13824.25 b.c.**

*Meaning:*

**The results of these calculations are negative numbers, indicating that the dates are before the time of Christ (BC) in the historic notation. The two dates from these calculations, 1155**

**b.c.**and 13824

**b.c.**are the reference dates in our past that tell us where to look in the existing cycles for the conditions that we can expect to repeat in 2012. The results of this comparison are summarized in Chapter 6, Figure 15.