Astronomy Explained Upon Sir Isaac Newton's Principles / And made easy to those who have not studied mathematics

by James Ferguson

To find the time of New or Full Moon in any given year and month before the Christian Æra.

356. Precept I. Find a year of the 18th Century, which added to the given number of years before Christ, diminished by one, shall make a number of whole Centuries.

II. Find this number of Centuries in [Table V], and subtract the Time and Anomalies answering to it from the Time and Anomalies answering to the mean New or Full Moon in March the year of the 18th Century thus found; and they will give the mean time of New or Full Moon in March the given year before Christ, with the Anomalies answering thereto. Whence the true time of that New or Full Moon may be had by the Precepts already delivered § [355].

III. The Tables are calculated for the Meridian of London: therefore, in computing for any place westward of London, four minutes of time must be subtracted from the time shewn by the Tables, for every degree the place is westward; and added for every degree it is eastward. See § [210].

EXAMPLE I.
To find the time of New Moon at London and Athens in March, the year before Christ 424.

The years 423 added to 1777 make 2200, or 22 Centuries.

	Days	Hours	Min.	Sun’s Anom.			Moon’s Ano.
				s	°	ʹ	s	°	ʹ
Tab. I. Mean New Moon in March A. D. 1777	27	7	53	9	7	27	5	25	51
From which subtract 2200 years in Tab. V.	6	21	47	11	16	26	4	20	37
Mean Conj. and Anom. in March before Chr. 424	20	10	6	9	21	1	1	5	14
Which with, the total of the three Equations added		9	20	Equ. Moon’s Anom.				-	19
				Anom. cor.			1	4	55
Gives the true time of Conjunction	20	19	26	Sun’s Equat.			+	1	48
Which was the 21st day of March, at 26 minutes past 7 in the morning at London: and if 1 hour 35 minutes be added for Athens, which is 23° 52ʹ east of the meridian of London, we have the time at Athens; namely, 1 minute past 9 in the morning.				Moon’s Ano.			1	6	43
				Moon’s ann. Eq.			0h	20m	add
				Her ellipt. Equ.			5	43	add
				Sun’s Equation			3	17	add
				Total			9	20	add

EXAMPLE II.
To find the time of Full Moon in October, the year before Christ 4030.

The years 1771 added to 4029 make 5800, or 58 Centuries.

			Days	Hours	Min.	Sun’s Anom.			Moon’s Ano.
						s	°	ʹ	s	°	ʹ
Tab. III. From the mean Full Moon in March 1771			19	7	11	8	29	6	7	22	30
Tab. V. Subtr. the numbers for 5800 years		5000	10	7	56	10	23	56	0	17	36
		800	5	4	43	11	27	43	7	7	7
Which collected make			15	12	39	10	21	39	7	24	43
Rem. the mean Full Moon &c. March before Chr. 4030			3	18	32	10	7	27	11	27	47
To which add eight Lunations to carry it to October			236	5	52	7	22	50	6	26	32
And the several sums will be			240	0	24	6	0	17	6	24	19
Which, for Full Moon day, Tab. VII, is October 26			26	0	24				h.	m.
Moon’s ellipt. Equation subtr. there being none besides				3	28	Moon’s Ann. Eq.			0	0	add
						Moon’s ellipt. Eq.			3	28	sub.
Rem. the true time of Full Moon, October			25	20	56	Sun’s Equation			0	0	add
Which is the 26th day, at 8 hours 26 minutes in the forenoon[[79]].						Total			3	28	sub.