Page:The Burmese & Arakanese calendars (IA burmesearakanese00irwiiala).pdf/44

The length of a mean lunation is, according to

Ball	29.530589	days
Young	29.530588	days„
Surya Siddhanta	29.530587946	days„

If these three be multiplied by 12863, the results are respectively

	379851.966307	days
	379851.953444	days„
and	379851.952749398	days„

Comparing the greatest estimate of 12863 mean lunations and the smallest estimate of 1040 mean tropical years,

12863 mean lunations	=	379851.966307	days
1040 mean tropical years	=	379851.879368	days„
the difference is only		000000.086939	days„

In 10,000 years the difference would amount to about 20 hours. This is a maximum estimate.

118. The best way to apply the cycle of 1040 years is to use it to make corrections in the Metonic cycle at regular intervals. The problem is to find at what intervals the correction should be made.

119. The number of watat in 1040 years is found by subtracting the solar from the lunar months.

12863 − 12480 = 383.

Now multiply the two kinds of cycles together.

19 × 1040 = 19760.

The number of watat in 19760 years is

By Meto	7	×	1040	=	7280
By de Cheseaux	383	×	19	=	7277
Difference					3

The Metonic cycle must be so modified as to cut out 3 watat in 19760 years.

120. The intervals between watat run in a series thus 3 3 3 2 3 3 2 and so on, over again. In each Metonic cycle there are two two-year intervals, one of which follows three three-year intervals, and the other follows two three-year intervals. Every correction must be made by converting into a three-year interval one of those two-year intervals which follow two three-year intervals

Thus

1	.	1	.	.	1	.	.	1	.	1	.	.	1	.	.	1	.	.
1	.	1	.	.	1	.	.	1	.	.	1	.	1	.	.	1	.	.
1	.	1	.	.	1	.	.	1	.	1	.	.	1	.	.	1	.	.
1	.	1	.	.	1	.	.	1	.	.	1	.	1	.	.	1	.	.
1	.	1	.	.	1	.	.	1	.	1	.	.	1	.	.	1	.	.		1
1	.	.	1	.	1	.	.	1	.	.	1	.	1	.	.	1	.	.		1