Find Ordinal Position of a Year in Chinese Lunar Calendar 60-Year Cycle

Chinese paintings and calligraphy traditionally use Tiangan (TG, literally "heaven stems") and Dizhi 
(DZ, literally "earthly branches") sequences to mark the date of the art work, e.g., "by <artist's 
name> at <place name> in the autumn month of the year Bing-Wu". The following calculation can 
convert TG-DZ year to the Gregorian year. This simple algorithm makes mental math possible and 
obviates the need to consult the conversion table.

TG and DZ are simply two sequences of numbers. There're ten in TG, named Jia, Yi,..., Gui, and 
twelve in DZ, named Zi, Chou,..., Hai. TG-DZ means one number in TG is paired with one in DZ, each 
incrementing and wrapping independently (see step (1) below); for example, Jia-Zi, Yi-Chou, ..., 
Gui-Hai. The pairing of all numbers in the two sequences forms 60 pairs, hence a lunar calendar 60-
year cycle, a.k.a. Huajia. If we know the Gregorian year for the Jia-Zi year (the first year) of a 
cycle, given the TG-DZ designation of any year in this cycle, we can find its Gregorian year as well. 
This note presents a mental math method for this calculation, but its proof has yet to come.[2022 Update]

(1) Convert TG-DZ to numbers. In the TG sequence, Jia, Yi, Bing, Ding, Wu, Ji, Geng, Xin, Ren, Gui 
are rewritten as 1, 2, ..., 10, respectively. In the DZ sequence, Zi, Chou, Yin, Mao, Chen, Si, Wu, 
Wei, Shen, You, Xu, Hai are rewritten as 1, 2, ..., 12, respectively. For instance, the year of 
Ji-Mao is recorded as (6,4), year of Bing-Wu is (3,7). The sixty pairs of TG-DZ are (1,1), (2,2), 
..., (10,10), (1,11), (2,12), (3,1), (4,2), ..., (9,11), (10,12).

For convenience, let's call the TG number (ordinal position in TG) a and the DZ number b.

(2) Calculate the ordinal position of this year in the 60-year cycle (Huajia). The unit digit is the 
a (see step (1)), e.g. 6 for year Ji-Mao (6,4), 3 for Bing-Wu (3,7).

(3) The tenth digit of the Huajia ordinal position is given as follows:

(a - b) / 2     if a >= b
(a - b) / 2 + 6 if a < b

For instance, the tenth digit of year Ji-Mao (6,4) is (6-4)/2=1. That of year Bing-Wu (3,7) is 
(3-7)/2+6=4.

(4) You must remember that year 1924 is a Jia-Zi (1,1) year; in fact, any year matching (1924+60*n), 
where n is an integer or 0, is a Jia-Zi. Thus, 1923 is the last year of the previous cycle (Gui-Hai 
or (10,12)). The Gregorian year for Ji-Mao (6,4) is 1923 + 16 = 1939, or 1983 + 16 = 1999 (1983 is 
also Gui-Hai); 16 is formed by the unit digit 6 (step (2)) and the tenth digit 1 (step (3)). Whether 
year Ji-Mao is 1939 or 1999 or even 739 A.D. is determined according to the context and other 
information. Similarly, year Bing-Wu (3,7) is 1923 + 43 = 1966 or 1966+60*n.

(5) If TG is Gui (the 10th number), e.g. year Gui-Chou (10,2), subtract 1 from both TG and DZ; e.g., 
Gui-Chou is reduced to Ren-Zi (9,1). Then calculate by following steps (1) through (4) and the 
Gregorian year is found to be 1923 + 49 = 1972 (let's assume it's in the 20th century). Lastly, add 
1 since we reduced Gui-Chou to Ren-Zi. So we know year Gui-Chou in this century is 1973.

Obviously, years using the TG-DZ notation are used by the "old" or lunar calendar. In fact, we cannot 
say for certain that years Jia-Zi of this century are 1924 and 1984, because it still depends on the 
month. In other words, we can say that the period after the Spring Festival of 1984 and before the 
Spring Festival of 1985 is Jia-Zi. The time before the 1984 Spring Festival is still Gui-Hai.

written in 1998
Email: yong321@yahoo.com
The Chinese version of this article is at
http://yong321.freeshell.org/misc/tiangandizhi.txt

[2022-10-25 Update] HL and JH have given the following explanation:

When a > b, this is when both knobs (a and b) have made the same number of turns.
Suppose n = the number of turns, you have
10n+a=12n+b therefore n=(a-b)/2
When a < b, the first knob has made one more turn than the second knob, so you have
10n+a=12(n-1)+b therefore n=(a+b)/2+6
Note: The first knob must turn either 0 or 1 more time than the second knob;
i.e. a turns either the same number of times as b, or 1 more time than b. If a > b, it has to be 0
(they turn the same number of times), because 10n+a=12n+b if and only if a>b. If a were to turn 2 more
times than b, the equation would be 10n+a=12(n-2)+b therefore n=(a-b)/2+12, which means b is at least 6.