Prophecy Kryptos: Prophetic Method With Mathematics Confirmation

      Welcome to "Prophecy Kryptos: Prophetic Method With Mathematics Confirmation," an Accompanying Crystalization best read with Website Prophecy "Prophecy Kryptos: Solving Kryptos K4, Tongues Confirmation, and Practical Persuasion."  Suggestion: begin with Part 1. Solving Kryptos K4.  The analysis of Kryptos K4 (the part 4 coding) is a demonstration of what we can achieve when we simply play mathematically and verbally with letters.  On this page's "Prophecy Kryptos: Prophetic Method With Mathematics Confirmation," prophecy and mathematics mix on the world map.

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Part 1. Observing the Apex and Berlin on the World Map Prior to Confirming Kryptos K4 Solved

1.a. Assessing the Apex and Berlin Clues Given to Solve Kryptos K4

      As he addressed questions about solving K4 (the part 4 coding) on the sculpture "Kryptos," sculptor James Sanborn would give different clues, among these "the apex of the pyramid" and "Berlin."  Yet, it was not until 10-20-12 that I decided to investigate in my atlas what I never knew existed: several cities Berlin in the U.S.  From my solution to K4 of BROOKLYNOK, I decided to make a two-dimensional triangle on the world map with Brooklyn, New York, its apex; Berlin, New Hampshire, its northernmost angle; and Langley, Virginia, its southernmost angle.  This time I hoped to more strongly confirm on the world map in the context of the prophetic method, using mathematics, my birthplace and K4 solution, Brooklyn.  Observe: my solution to K4.

1.b. Creating on the World Map the Triangle Whose Apex Is Opposite a Side With a Test Midpoint

      Point A = (38.57 N, 77.80 W) = Kryptos in Langley, VA                                 (degree signs are omitted)
      Point B = (40.65 N, 73.95 W) = Apex clue in Brooklyn, NY
      Point C = (44.47 N, 71.19 W) = Berlin clue in Berlin, NH
      Point D = (41.52 N, 74.50 W) = Midpoint between Point A and Point C

      Point A is part of the solution to K2 (its contrived value there of degrees, minutes, and seconds changed to its contrived value here of a decimal number), while Point B and Point C are estimates of coordinates from Google.  Point D is calculated from (38.57 + 44.47)/2 = 41.52 and (77.80 + 71.19)/2 = 74.50.  Notice that when adding the coordinates of each point of the triangle, Point A = 38.57 + 77.80 = 116.37 near 116, Point B = 40.65 + 73.95 = 114.60 near 115, and Point C = 44.47 + 71.19 = 115.66 near 116.  Interestingly, the total for Point B = Apex clue in Brooklyn, NY, of 115 is 1 lower than the total of 116 for each of the other 2 points, while the total for "BERLIN" decoded from the 64th through 69th letters of K4 = NYPVTT = 14 + 25 + 16 + 22 + 20 + 20 = 117 is 1 higher than the total of 116 for each of the other 2 points.  Also, NY = 14 + 25 = 39, and 38.57 is near 39, while PVTT = 16 + 22 + 20 + 20 = 78, and 77.80 is near 78.  Finally, BROOKLYN = 2 + 18 + 15 + 15 + 11 + 12 + 25 + 14 = 112, exactly 4 away from 116, 4 being the number commonly occurring in the solution to K4.

      Following is a sketchy but ample graph of the points on the triangle on the world map with Point S for Hurricane Sandy:

45__                                                                                              
                                                                            C o 
44__

43__
                                                                              
42__
                                   D
                                               
o
41__                                          
                                                    o
B
40__                                           

39__
           A o 
              
38__                                                    
o S

             |         |         |         |         |         |         |         |
           78      77      76     75      74      73     72      71

      Notice that observation does not establish Point B as the exact apex of triangle ABC, because triangle ABC is not an isosceles triangle: Point B is not equidistant to Point A and to Point C, and a line drawn from Point B to Point D does not form two right angles at Point D, which is the midpoint of the base of triangle ABC.  Indeed, we can use mathematics to demonstrate these facts with rigor, and we have no intention of disregarding mathematical rigor.  However, our aim is to demonstrate that Point B = (40.65 N, 73.95 W) = Apex clue in Brooklyn, NY, confirms the solution to K4 of BROOKLYNOK.  We were given an apex clue.  Therefore, can we demonstrate that although Point B is not the exact apex of triangle ABC, there is a high probability that Point B in Brooklyn, NY, confirms the solution to K4 of BROOKLYNOK?   

Caution about slope (slant): 1) The coordinates of points on the world map are given in opposite positions: unlike (x,y) on a graph, they are (y,x).  Therefore, when determining the slope of a line on the world map, we have to switch the positions of the coordinates.  For slope associated with the above map, we must put the N coordinates running south-north in the numerators and the W coordinates running east-west in the denominators.  2) Since the above W coordinates are runnning east-west instead of west-east, at some point in our slope calculations, we must also change the sign of the W coordinates to reflect the opposite direction. 

Part 2. Using the Prophetic Method With Mathematics For Confirming Kryptos K4 Solved

2.a. Developing The Kryptos K4 Revelation of <90% to Practice the Prophetic Method

      We can demonstrate that although Point B is not the exact apex of triangle ABC, there is a high probability that Point B in Brooklyn, NY, confirms the solution to K4 of BROOKLYNOK by using the prophetic method with mathematics.  The prophetic method yields a revelation, or a word of wisdom, which consists of three or four words of knowledge.  Of these words of knowledge, the first is usually discarded and the last is a sign following.  Following is The Kryptos K4 Revelation of <90% to practice the prophetic method.

      1. The sculpture "Kryptos" was dedicated in '90 = 90.  In prophecy, year 90 can be 90% (or =90%), because we read in The Holy Bible that "And he changeth the times [from year to %] and the seasons: he removeth kings, and setteth up kings: he giveth wisdom unto the wise, and knowledge to them that know understanding:" (Dan. 2.21).  Yet, we must discard the word of knowledge that we have =90%, because '90 = 1990, while the remaining words of knowledge yield <90%.     

      2. In triangle ABC, we can calculate the slope from Point A to Point D (half the triangle's base) as follows:

38.57 - 41.52 = -2.95      -2.95 = roughly .89 = 89%      89% <90% 
77.80 - 74.50      3.30      -3.30

      3. In triangle ABC, we can calculate the slope from Point C to Point D (half the triangle's base) as follows:

44.47 - 41.52 =  2.95      2.95 = roughly .89 = 89%      89% <90% 
71.19 - 74.50    -3.31      3.31

      4. In triangle ABC, we can calculate the slope from Point A to Point C (the triangle's base) as follows: 

38.57 - 44.47 = -5.90      -5.90 = roughly .89 = 89%      89% <90% 
77.80 - 71.19      6.61      -6.61

We expected this word of knowledge to be a sign following from the two words of knowledge before it.

      Analyzing all four words of knowledge--the revelation--we have failed to demonstrate the high probability we sought: not only are the retained words of knowledge <90%, none of their calculations include Point B. 

2.b. Developing The Kryptos K4 Revelation of =90% or >90% to Achieve the Statistical Significance

      At last, we can demonstrate that although Point B is not the exact apex of triangle ABC, there is a high probability that Point B in Brooklyn, NY, confirms the solution to K4 of BROOKLYNOK by using the prophetic method with mathematics.  Following is The Kryptos K4 Revelation of =90% or >90% to achieve the statistical significance. 

      1.  Recall that I mentioned that we can both observe and demonstrate mathematically that a line drawn from Point B to Point D does not form two right angles at Point D, which is the midpoint of the base of triangle ABC.  Hence we need to investigate the line which would form two right angles with the base of triangle ABC, although not at Point D: the perpendicular line to the base of triangle ABC.  By definition, this perpendicular line from Point B to some unknown Point E on the base of triangle ABC must have a slope which is the negative reciprocal of the slope of the base of triangle ABC.  We already know that this slope from Point A to Point C = .89.  Therefore, the slope of the perpendicular line from Point B to some unknown Point E on the base of triangle ABC must be -1/.89 = roughly -1.12.  Now, contrast -1.12 with the slope on the above map from Point B to Point D:

40.65 - 41.52 = -.87      -.87 = roughly -1.58
73.95 - 74.50    -.55       .55

Since the slope of -1.58 from Point B to Point D is mathematically so far away from the slope of -1.12 of the perpendicular line from Point B to an unknown Point E, this word of knowledge does not yield a finding of statistical significance (except <90%) so that we must discard this word of knowledge.  Nevertheless, this is not the time to abandon our aim, the prophetic method with mathematics, the prophetic method, or mathematics, as shown by the following words of knowledge.       

      2. In triangle ABC, we can calculate the distance from Point B to Point D as follows:

(40.65 - 41.52)2 + (73.95 - 74.50)2 = (-.87)2 + (-.55)2 = .76 + .30 = 1.06      square root of 1.06 = 1.03 = roughly 1

This word of knowledge informs us that on the above map Point B is roughly 1 degree away from Point D.  Considering the size of the world map in its entirety, this finding revives the "promise" that there is a high probability that Point B in Brooklyn, NY, confirms the solution to K4 of BROOKLYNOK.  We can give this finding a conservative statistical significance of =90%.  If you are hoping that the next word of knowledge fulfills the "promise," know that I am, too. 

      3. Our daunting challenge is to discover a liberal statistical significance of >90%.  I decided that what I needed to do so was more faith, and I remembered its definition, "Now faith is the substance of things hoped for, the evidence of things not seen" (Heb. 11.1).  That's my answer: the clue that I now need is unseen.  Then, I wondered, "What is the slope of the unseen line passing through unknown Point E and perpendicular to the line from Point B to Point D at some unknown Point F?  In other words, assume that Point B is correct and seesaw at unknown Point E the line loosed from Point A and from Point C until it is perpendicular to the line from Point B to Point D at some unknown Point F.  We already know that the slope of the line from Point B to Point D = -1.58.  The slope of the unseen line passing through unknown Point E and perpendicular to the line from Point B to Point D at unknown Point F must be the negative reciprocal of -1.58, or -1/-1.58 = .63.  Compare this slope (which is the slope of the unseen seesawing line at unknown Point E) to the slope of the line from Point A to Point C = .89 as follows:

.89 - .63.26 = roughly .29 = 29%      100% - 29% = 71%      Slope compression lands us closer to Point B.
     .89        .89

What we have shown is that Point D is farther from Point B than unknown Point F is closer to Point B.  If we draw a circle whose center is at Point B and whose circumference passes through Point D, its radius would be as we have already shown: the distance from Point B to Point D = roughly 1 degree.  However, if we draw a circle whose center is at Point B and whose circumference passes through unknown Point F and unknown Point E, its radius (to either Point F or Point E) is substantially shorter than the roughly 1 degree radius.  The 71% might have another mathematical application.  Such is for the mathematician to determine.  This finding fulfills the "promise" that there is a high probability that Point B in Brooklyn, NY, confirms the solution to K4 of BROOKLYNOK.  We can give this finding a liberal statistical significance of >90%.

      4.  I demonstrated =90% alone and >90% alone.  Now, what sign following can demonstrate =90% and >90% together?  I completed the other pages of "Prophecy Kryptos: Solving Kryptos K4, Tongues Confirmation, and Practical Persuasion" on 07-27-12, and today is 10-25-12. 

      Counting from 07-27-12 through 10-25-12 and omitting 07-27-12 yields 90 days = 90%
      Counting from 07-27-12 through 10-25-12 and including 07-27-12 yields 91 days = 91%

This word of knowledge yields one finding of statistical significance of =90% and another finding of statistical significance of >90%.

      Ordinarily, such a word of knowledge would be regarded as debatable and merit discarding.  However, when you wonder why we even need the prophetic method, realize that as soon as I got the realization of the upcoming 90 days or 91 days, this understanding inspired me to produce the two revelations here.  I do not believe that science alone can accomplish as much in so short a period of time, because as already cited, ". . . he giveth wisdom unto the wise, and knowledge to them who know understanding:" (Dan. 2.21).  Having begun to think about this page on 10-20-12, I finished all of the work associated with Part 1 and Part 2 on 10-25-12 at 11:52 pm, and I whizzed this work onto the Internet.  (I admit that I corrected the third word of knowledge on 11-01-12.  I should have heeded my hunch to discuss circles, because all hunches are words of knowledge.  Oops, I also admit that I again corrected the third word of knowledge on 01-27-13.  I would not have been so continually sloppy had the cares of the world not been so distracting.  From 07-27-12 to 01-27-13 is 6 months and 1 day, the 1 day here being interpreted as 1 God.)

Part 3. Peeking at the Hand of God in the Act of God Hurricane Sandy

3.a. The Slope of the Line From Point B to Unknown Point E Appears Again as an Estimation Within Triangle ABC  

       We did not need Hurricane Sandy to find an estimation within triangle ABC of the slope of the line from Point B to unknown Point E, which is the perpendicular line to the base of triangle ABC.  We already know that the slope of the perpendicular line = -1.12, but we will use -1.1 (because Hurricane Sandy's coordinates are given to one decimal place).  We seek the estimation of -1.1 by comparing distances within triangle ABC.

      Within triangle ABC are triangle ABD and triangle CBD.  These subtriangles share the line from Point B to Point D.  Also, the distance from Point A to Point D = the distance from Point C to Point D = 4.5.  Therefore, remaining are--and we are seeking their ratio--the distance from Point C to Point B/the distance from Point A to Point B = 4.7/4.3 = 1.1, and while distance has no sign, make this -1.1 to reflect the W coordinates running east-west.  So we have -1.1 and again -1.1.  We calculate distances (to one decimal place) as follows:

1/2 the distance from Point A to Point C: (38.6 - 44.5)2 + (77.8 - 71.2)2 = (-5.9)2 + (6.6)2 = 34.8 + 43.6 = 78.4      square root of 78.4 = 8.9      1/2 x 8.9 = 4.5

the distance from Point C to Point B: (44.5 - 40.7)2 + (71.2 - 74.0)2 = (3.8)2 + (-2.8)2 = 14.4 + 7.8 = 22.2      square root of 22.2 = 4.7

the distance from Point A to Point B: (38.6 - 40.7)2 + (77.8 - 74.0)2 = (-2.1)2 + (3.8)2 = 4.4 + 14.4 = 18.8      square root of 18.8 = 4.3        

3.b. The Slope of the Line From Point B to Point D Appears Again as an Estimation Within Quadrilateral ASCB

       I had first completed Part 1 and Part 2 4 days prior to 10-29-12 when CNN's Wolf Blitzer of "The Situation Room," beginning at 4:00 pm, was covering Hurricane Sandy, located on a screen banner at (38.3 N, 73.1 W).  See this Point S = (38.3 N, 73.1 W) added to the above map.  Now, we can find an estimation within quadrilateral ASCB (Point B is not on the line from Point A to Point C) of the slope of the line from Point B to Point D, which we already know is -1.58, but we will use -1.6 (because Hurricane Sandy's coordinates are given to one decimal place).  We seek the estimation of -1.6 by comparing distances within quadrilateral ASCB.

      Within quadrilateral ASCB are triangle ASB and triangle CSB.  These subtriangles share the line from Point S to Point B.  Also, it seems that the distance from Point A to Point S = the distance from Point C to Point B = 4.7.  Therefore, remaining are--and we are seeking their ratio--the distance from Point C to Point S/the distance from Point A to Point B = 6.5/4.3 = 1.5, and while distance has no sign, make this -1.5 to reflect the W coordinates running east-west.  So we have -1.6 and now -1.5.  We calculate distances (to one decimal place) as follows:

the distance from Point A to Point S: (38.6 - 38.3)2 + (77.8 - 73.1)2 = (.3)2 + (4.7)2 = .1 + 22.1 = 22.2      square root of 22.2 = 4.7

the distance from Point C to Point B: (44.5 - 40.7)2 = (71.2 - 74.0)2 = (3.8)2 + (-2.8)2 = 14.4 + 7.8 = 22.2      square root of 22.2 = 4.7

the distance from Point C to Point S: (44.5 - 38.3)2 + (71.2 - 73.1)2 = (6.2)2 + (-1.9)2 = 38.4 + 3.6 = 42.0      square root of 42.0 = 6.5

the distance from Point A to Point B: (38.6 - 40.7)2 + (77.8 - 74.0)2 = (-2.1)2 + (3.8)2 = 4.4 + 14.4 = 18.8      square root of 18.8 = 4.3

      Compare -1.6 and -1.1 and find that (-1.6) - (-1.1) = -.5.  Compare -1.5 and -1.1 and find that (-1.5) - (-1.1) = -.4.  Compare -1.6 and -1.5 and find that (-1.6) - (-1.5) = -.1.  Here, the negation stops us from concluding that point 5 and point 4 and point 1 = points 5 and 4 and 1 = point's 5 and 4 and 1 = point is 5 and 4 and 1, meaning that 5 from 4 and 1 repeat, as in the steps to the solution to K4.  In other words, these demonstrated Hurricane Sandy relationships might be mathematically repetitive relationships or nonmathematically coincidental relationships common among 5 triangles and not the result of divine intervention.  Such is for the mathematician to determine.  The 5 triangles can be the 4 basic adjacent triangles (ABD, CBD, ABS, and CBS) within 1 triangle (ASC).  Or, the 5 triangles can be the 4 basic adjacent triangles (ABD, CBD, ABS, and CBS) turned into 1 more triangle (ABD turned into 2 triangles with the addition of unknown Point E) for a total of 5 basic adjacent triangles (ABE, DBE, CBD, ABS, and CBS).  We do have 5 from 4 and 1 repeating 2 times, as in the steps to the solution to K4.  Yet, the astute thinker knows that we also do have >5 triangles when we seesaw the unseen line perpendicular to the line from Point B to Point D.  Thus, realize that like the passing winds and rain of a disfiguring hurricane, we have rescued but a brief glimpse of this work's numbers commonly occurring: 5, 4, and 1.  If you believe that a hurricane is an act of God, then you already believe that it has the hand of God in it.  True prophecy tells the truth, and the true prophet or the true prophetess, by the leading of the Holy Spirit (God's Spirit), uses the prophetic method to try to attain the truth.  Do you think that these estimations on the flat world map are equalities on the curved globe?  If we could show such equalities, then we would have proof of the existence of an intelligence other than ours (God), would we not?  Having added to this page Hurricane Sandy of 10-29-12, I finished all of the work associated with Part 3 on 10-30-12 at 11:56 pm, and I whizzed this work onto the Internet.  (I admit that I carefully reworded this paragraph on 11-01-12.  I also admit that I again carefully reworded this paragraph and even the one before it on 01-27-13.  From 10-20-12 to 01-27-13 is 100 days.)

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Carolyn Jane Siino is the copyright owner of all content (except the Kryptos coding) described on www.awakenedtodoprophecy.com
page content: online 10-25-12 (Part 1 and Part 2) and online 10-30-12 (Part 3), last revised 01-28-13