Page:On Chronology and the Construction of the Calendar.pdf/37

happened

	Mean Peking time
the entrance of the sun in the sign Yeo of the zodiac	April 20 7h 12m P.M.
Newmoon	April 21 9h 41m P.M.
Newmoon	May 21 1h 36m P.M.
the entrance of the sun in the sign Shen of the zodiac	May 21 7h 06m P.M.

During the Lunation April 21 9^h 41 P.M. until May 21 1^h 36^m P.M. the sun did not enter in a new sign of the zodiac; it was during the whole Lunation in the sign Yeo and had not passed over an even tsie-khi, but only over the odd tsie-khi Li Hia, the 7^th tsie-khi of our table (25).

For this reason, the Lunation April 28 until May 20 in the Chinese calendar for the Chinese year, commencing in 1879 A.D., received the N^o3, the same as the antecedent month and was the intercalary month.

The standard scale of the Chinese Chronology is the cycle of 60 years, similar to the Julian period of 1980 years.

The first year of the first sexagesimal cycle was the year 2637 B.C. or −2637, and the 57^th year of the 44^th cycle corresponds to −1 or 1 B.C.; the 58^th year of the 44^th cycle corresponds to +1 or 1 A.D. and the first year of the 76^th cycle to +1864 or 1864 A.D.

11. Comparison between Chinese and European chronological dates.

As the Chinese Newyear occurs between the 21 January and 20 February the Chinese twelfth month occurs always partly, sometimes entirely in the next European year; likewise the Chinese eleventh month sometimes extends into the next European year. It further follows, that the N^o of the European month, if January has N^o1,February N^o2 etc. is mostly by 1, sometimes by 2 greater than the N^o of the Chinese month, sometimes equal to that of the latter.

If given the Chinese sexagesimal cycle $c$ and the year $y$ of this cycle and if we require the year $T$ of the Christian era, which has in common with $y$ at least 10 months, we find,

when $T$ negative or for all time before and equal to the 57^th year of the 44^th cycle

Equation (26)

$T=c{\,.\,}60+y-2698$

and, when $T$ positive or for all time after and equal to the 58^th year of the 44^th cycle,

Equation (27)

$T=c{\,.\,}60+y-2697$

Vice versa, when $T$ given and required $c$ and $y$, we get, if $T$ negative,

Equation (27)

$T+2698=c{\,.\,}60+y$

and the rule to find $c$ and $y$ is thus: if 60 in $T+2698$ is contained $n$ times and the remainder is $r$, we have the identical equations

Equation (27)

$T+2698=n{\,.\,}60+r=c{\,.\,}60+y$

Equation (28)

therefore the required $c=n$

Equation (29)

and the required„ $y=r$

When $T$ positive, we obtain from equation (27):