BTC _ FINDING ACCESS CODE TO MATRIX (Orange Path) Update 1
Mis à jour
Next ATL : 22.02.2020
Next ATH: 06.02.2021
Second ATL: 01.03.2025
This is Not A Financial Advice, It is A Study Case For Myself.
I will try to share the development, as simple as I can ;
DAY ONE (on chart) to ATH-1 , we have 325 days (included). So I tried to find a sequence with fibonacci. What I found was ATH-1 to ATH-2 is following 2,618 fibonacci days with some errors, and ATH-2 to ATH-3 1,618 fibonnaci days with some errors. Actual days were different than what is expected with fibonacci number sequences. So digged and tried to find the lateness (gap).
.................(sequence/1,618) =........................ 2,618 / 1,618=1,618 .......................1,618 / 1,618=1
ATH0-ATH1...................... ATH1-ATH2 ................................ ATH2-ATH3 ....................................... ATH3-ATH4
325 .................... 325 * 2,618 = 850,85 .............850,85 * 1,618 = 1376,68..................so; 1376,68 * 1 = 1376,68 (?)
325..................................... 906 ..................................... 1478
..... 0............................. 906-850,15 = 55,15 days of error...............1478-1376,68= 145,77 days of error
same can be applied to bottoms ;
.................(sequence/1,618) =........................ 2,618 / 1,618=1,618 .......................1,618 / 1,618=1
ATL0-ATL1...................... ATL1-ATL2 ................................ ATL2-ATL3 ....................................... ATL3-ATL4
406 .................406 * 2,618 = 1062,91 .............1062,91 * 1,618 = 1719,79 .............if so; 1719,79 * 1 = 1719,79
406 .....................................1153 .................................. 1865,56 (?)................................. 1865,56 (?)
....................................1062,91-1153 = 90 days of error
How I found the next ATL and Next ATH (numer of days with question mark) ?
----------------------------------------------------------------------------------------------------------------------------
When I sum previous expected ATL with next expected ATL and divide it to error days, I found a fixed number which also aplied to next cycles.
(ATL0-ATL1 ) + (ATH1-ATH2 ) / 55,15 days of error
(406) + (850,85) /55,15 = 22,85
(ATL1-ATL2 ) + (ATH2-ATH3 ) / 101,32 days of error (5 days is lost in chartin this period)
(1062,91) + (1376,68) / 106 = 22,85 lateness secret ratio (silence...)
then next ATH should be ;
(ATL2-ATL3 ) + (ATH3-ATH4 )]
(1719,79) +(1376,68) / secret ratio of 22,85 = 112,46 days of delay.
So expected ATH2-ATH3 is 1376,68 days + 112,46 days = 1512 days = 06.08.2020
-------------------------------------------------------------------------------------------------------------------------------
Now bottom ;
(ATH0-ATH1 ) + (ATL0-ATL1 ) has no delays , we accepted them as realized as expected. (to create a base)
(ATH1-ATH2 ) + (ATL1-ATL2 ) / 90 days of error
(850,85) + (1062,91) /90 = 21,24
I assume that 21,24 also can be applied which above 22,85 is happened twice.
then next ATL should be ;
(ATH1-ATH2 ) + (ATL1-ATL2 ) / 21,24 lateness secret ratio
(1376,68) + (1719,79) /21,24 = 145,77 days of delay
So expected ATL2-ATL3 is 1719,79 days + 145,77 days = 1865 days = 22.02.2020
---------------------------------------------------------------------------------------------------------------
ATL3-ATL4 same fibonacci expected days and delayed days applied. Since the sequence will follow 2,618 & 1,618 & 1,000
ORANGE PATH (First Version)
BLUE PATH (Update 1)
BLUE PATH (First Version)
Next ATH: 06.02.2021
Second ATL: 01.03.2025
This is Not A Financial Advice, It is A Study Case For Myself.
I will try to share the development, as simple as I can ;
DAY ONE (on chart) to ATH-1 , we have 325 days (included). So I tried to find a sequence with fibonacci. What I found was ATH-1 to ATH-2 is following 2,618 fibonacci days with some errors, and ATH-2 to ATH-3 1,618 fibonnaci days with some errors. Actual days were different than what is expected with fibonacci number sequences. So digged and tried to find the lateness (gap).
.................(sequence/1,618) =........................ 2,618 / 1,618=1,618 .......................1,618 / 1,618=1
ATH0-ATH1...................... ATH1-ATH2 ................................ ATH2-ATH3 ....................................... ATH3-ATH4
325 .................... 325 * 2,618 = 850,85 .............850,85 * 1,618 = 1376,68..................so; 1376,68 * 1 = 1376,68 (?)
325..................................... 906 ..................................... 1478
..... 0............................. 906-850,15 = 55,15 days of error...............1478-1376,68= 145,77 days of error
same can be applied to bottoms ;
.................(sequence/1,618) =........................ 2,618 / 1,618=1,618 .......................1,618 / 1,618=1
ATL0-ATL1...................... ATL1-ATL2 ................................ ATL2-ATL3 ....................................... ATL3-ATL4
406 .................406 * 2,618 = 1062,91 .............1062,91 * 1,618 = 1719,79 .............if so; 1719,79 * 1 = 1719,79
406 .....................................1153 .................................. 1865,56 (?)................................. 1865,56 (?)
....................................1062,91-1153 = 90 days of error
How I found the next ATL and Next ATH (numer of days with question mark) ?
----------------------------------------------------------------------------------------------------------------------------
When I sum previous expected ATL with next expected ATL and divide it to error days, I found a fixed number which also aplied to next cycles.
(ATL0-ATL1 ) + (ATH1-ATH2 ) / 55,15 days of error
(406) + (850,85) /55,15 = 22,85
(ATL1-ATL2 ) + (ATH2-ATH3 ) / 101,32 days of error (5 days is lost in chartin this period)
(1062,91) + (1376,68) / 106 = 22,85 lateness secret ratio (silence...)
then next ATH should be ;
(ATL2-ATL3 ) + (ATH3-ATH4 )]
(1719,79) +(1376,68) / secret ratio of 22,85 = 112,46 days of delay.
So expected ATH2-ATH3 is 1376,68 days + 112,46 days = 1512 days = 06.08.2020
-------------------------------------------------------------------------------------------------------------------------------
Now bottom ;
(ATH0-ATH1 ) + (ATL0-ATL1 ) has no delays , we accepted them as realized as expected. (to create a base)
(ATH1-ATH2 ) + (ATL1-ATL2 ) / 90 days of error
(850,85) + (1062,91) /90 = 21,24
I assume that 21,24 also can be applied which above 22,85 is happened twice.
then next ATL should be ;
(ATH1-ATH2 ) + (ATL1-ATL2 ) / 21,24 lateness secret ratio
(1376,68) + (1719,79) /21,24 = 145,77 days of delay
So expected ATL2-ATL3 is 1719,79 days + 145,77 days = 1865 days = 22.02.2020
---------------------------------------------------------------------------------------------------------------
ATL3-ATL4 same fibonacci expected days and delayed days applied. Since the sequence will follow 2,618 & 1,618 & 1,000
ORANGE PATH (First Version)
BLUE PATH (Update 1)
BLUE PATH (First Version)
Note
explanation corrections without change in results ; ATH2-ATH3 = not 145,77 but 106 days of error
and
(ATL1-ATL2 ) + (ATH2-ATH3) / 22,85
(1719,79) +(1376,68) / secret ratio of 22,85 = 135,5 days of delay.
ATH3-ATH4 is 1376,68 days + 135,50 days = 1512 days = 06.08.2020
Note
One more correction; ATH3-ATH4 is 1376,68 days + 135,50 days = 1512 days = 06.02.2021
Ordre annulé
This idea is no more valid due to next ATL is not realized on 22nd Feb.
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