BTC _ FINDING ACCESS CODE TO MATRIX (Orange Path) Update 1
Обновлено
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)
Заметка
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
Заметка
One more correction; ATH3-ATH4 is 1376,68 days + 135,50 days = 1512 days = 06.02.2021
Заявка отменена
This idea is no more valid due to next ATL is not realized on 22nd Feb.
Похожие публикации
Отказ от ответственности
Все виды контента, которые вы можете увидеть на TradingView, не являются финансовыми, инвестиционными, торговыми или любыми другими рекомендациями. Мы не предоставляем советы по покупке и продаже активов. Подробнее — в Условиях использования TradingView.