//Your 1B, is extending over a two year period, going from 2B to 1A. //
Sorry, that is a formatting error, not a years error.
Here, is some stuff where years can be compared.
(the comparable stuff is on the same line
/-- with 0 year
/---------/----- without 0 year
/---------/---------/---- the AA method
/---------/---------/
4BC --- 4BC --- 4BC
3BC --- 3BC ---3BC
2BC --- 2BC --- 2BC
1BC --- 1BC --- AD1BC
-0- --- AD1 --- AD2
AD1 --- AD2 --- AD3
AD2 --- AD3 --- AD4
AD3 --- AD4 ----AD5
AD4 --- AD5 --- AD6
with 0 year shifts 0
no 0 yerars shifts 1 -- this is correct
AA's methoid shifts 2
// [You are pulling out a 0 mark on the time line, not a 0 year.] //
I'm doing it right.
I'm a retired computer professional engineer.
I know how to count with zero in and
without zero.
Honeywell numbers bits (old eight bit computer)
012345678 - '8' bit is the most significant figure
IBM numbered their bits (old 16 bit computer)
FEDCBA9876543210 -'0' bit is the most significant
(/which makes way more sense as the bit#
is the power of two of the bit
x to the 0 power = 1, if X is a counting number
and not equal to zero. Needless to say, the
IBM compatable computer came to be the
standard for the industry./)
Tandy numbered their bits (old eight bit computer)
123456789 - '9' bit is the most significant computer
Anybody ever program a computer with
'1's & '0's?
(it is slow. I love high order languages
over assembler languages over '1's & '0's.
What was the bit order over at Apple? It is the
only think I never messed with
BTW, 01 Jan AD0001 is one year after 01 Jan 0001BC
AD0001 runs from 01 Jan AD0001 to 31 Dec AD0001
0001BC runs from 01 Jan 0001BC to 31 Dec 0001BC
01 Jan AD0001 is the day after 31 Dec 0001BC
If there was a year zero (and some systems there are) it would:
start on 01 Jan 0000 (the day after 31 Dec 0001BC)
and end 31 Dec 0000 (the day before 01 Jan AD0001).