CHAP. XIX.
 
The Calculation of New and Full Moons and Eclipses. The geometrical Construction of Solar and Lunar Eclipses. The examination of antient Eclipses.

353. To construct an Eclipse of the Sun, we must collect these ten Elements or Requisites from the following Astronomical Tables.

Requisites for a solar Eclipse.

I. The true time of conjunction of the Sun and Moon: to know at what conjunctions the Sun must be eclipsed; and to the times of those conjunctions,

II. The Moon’s horizontal parallax, or angle which the semi-diameter of the Earth subtends as seen from the Moon.

III. The Sun’s true place, and distance from the solstitial colure to which he is then nearest, either in coming to it or going from it.

IV. The Sun’s declination.

V. The angle of the Moon’s visible path with the Ecliptic.

VI. The Moon’s Latitude or Declination from the Ecliptic.

VII. The Moon’s true hourly motion from the Sun.

VIII. The Angle of the Sun’s semi-diameter as seen from the Earth.

IX. The Angle of the Moon’s semi-diameter as seen from the Earth.

X. The semi-diameter of the Penumbra.

And for an Eclipse of the Moon, the following Elements.

Requisites for a lunar Eclipse.

I. The true time of opposition of the Sun and Moon; and for that time,

II. The Moon’s horizontal parallax.

III. The Sun’s semi-diameter.

IV. The semi-diameter of the Earth’s shadow.

V. The Moon’s semi-diameter.

VI. The Moon’s Latitude.

VII. The Moon’s true hourly motion from the Sun.

VIII. The Angle of the Moon’s visible path with the Ecliptic.

These Elements are easily found from the following Tables and Precepts, by the common Rules of Arithmetic.

Note, 60 minutes make a Degree, 30 degrees a Sign, and 12 Signs a Circle. A Sign is marked thus s, a Degree thus °, and a Minute thus ʹ.

When you exceed 12 Signs, always reject them and set down the remainder. When the number of Signs to be subtracted is greater than the number you subtract from, add 12 Signs to that which you subtract from; and then you will have a remainder to set down.

How the Signs are reckoned.

354. As we fix arbitrarily upon the beginning of the Sign Aries to reckon from, when we speak of the places of the Sun, Moon, and Nodes; we call Aries 0 Signs, Taurus 1 Sign, Gemini 2 Signs, Cancer 3 Signs, &c. So, when the Sun is in the 10th degree of Aries, we say his Place or Longitude is 0 Signs 10 Degrees, because he is only 10 Degrees from the beginning of Aries: if he is in the 5th, 10th, &c. Degree of Taurus, we say his Place or Longitude is 1 Sign, 5, 10, &c. Degrees: and so on, till he comes quite round again. But in reckoning the Anomalies of the Sun and Moon, and their distance from the Nodes, we only consider the number of Signs and Degrees the Luminaries are gone past their Apogee or Nodes; not how far they have to go to these points, were the distance ever so little. The Sun, Moon, and Apogee move according to the order of Signs, but the Nodes contrary. We shall now give the Precepts and Examples for the above Requisites in their due order.

To calculate the time of New and Full Moon.

First Element or Requisite.

355. Precept I. For any proposed year in the 18th Century, take out the mean time of the New Moon in March from Table I., and the mean time of Full Moon from Table III., for the Old Stile; or from Tables II and IV for New Stile; with the mean Anomalies of the Sun and Moon for these times, and set them by themselves. Then, from Table VI, take out as many Lunations as the proposed Month is after March, with the days, hours, and minutes belonging to them; and also the mean Anomalies of the Sun and Moon for these Lunations.

II. Add the days, hours, and minutes of these Lunations to the time of New or Full Moon in March, and the Anomalies for the Lunations to the Anomalies for March: the sums give the hours and minutes of the mean New or Full Moon required, and the mean Anomalies of the Sun and Moon for that time.

III. Then, with the number of days enter Table VII, under the given Month, and right against this number, in the left hand column you have the day of New or Full Moon; which set before the hours and minutes above-mentioned.

IV. But, (as it will sometimes happen) if the number of days fall short of all those under the given Month, add one Lunation with its Anomalies from Table VI to the foresaid sums; so you will have a new sum of days wherewith to enter the 7th Table under the given Month, where you are sure to find that sum the second time, if the first falls short.

V. With the Signs and Degrees of the Sun’s Anomaly enter Table VIII, The Moon’s annual Equation, and take out the minutes of time of that Equation by the Anomaly; remembring, that if the Signs are at the head of the Table, the degrees are at the left hand, in which case the Equation found in the Angle of meeting must be subtracted from the mean time of New or Full Moon, as the title Subtract, at the head of the Table directs: but if the Signs are at the foot of the Table their degrees are in the right-hand column, and the Equation where the Signs and Degrees meet in the Table is to be added to the mean time, as the title Add, at the foot of the Table directs; which Equation, so applied, gives the mean time of New or Full Moon corrected.

VI. With the Signs and Degrees of the Sun’s Anomaly enter Table IX, Equation of the Moon’s mean Anomaly, and take out the Equation thereof; adding it to the mean Anomaly or subtracting it therefrom, as the titles at the head or foot of the Table direct; and it gives the mean Anomaly corrected. Then, with the Sun’s Anomaly enter Table XII, Equation of the Sun’s mean Place, and take out that Equation, applying it to the Moon’s corrected Anomaly as the titles direct; and it will give the Moon’s Anomaly equated[77]. N. B. In all these Equations, care must be taken to make proper allowance for the odd minutes of Anomaly; the Tables having the Equations only for compleat Degrees.

VII. With the Moon’s equated Anomaly enter Table X, The Moon’s elliptic Equation, and take out that Equation in the same manner as the preceding: adding it to the former corrected time if the Signs be at the head of the Table, or subtracting it if they be at the foot, as the Table directs; and this gives the mean time equated.

VIII. Lastly, enter Table XI, The Sun’s Equation at New and Full Moon, with the Sun’s Anomaly, and take out the Sun’s Equation in the same manner as the others; adding it to, or subtracting it from the former equated time, as the titles direct: and by this last Equation you have the true time of New or Full Moon, agreeing with well regulated Clocks and Watches. But to make it agree with true Sun-Dials, the Equation of time must be applied as taught § 225.

EXAMPLE I.
 
To find the time of New Moon in April 1764, N. S.

  Days Hours Min. Sun’s Anom. Moon’s Ano.
s ° ʹ s ° ʹ
Tab. II. Mean time of New Moon in March 2 8 57 8 2 23 10 13 32
Add, for Lunation, from Tab. VI. 29 12 44 0 29 6 0 25 49
Mean New Moon and Anomaly 31 21 41 0 1 29 11 9 21
To which Time add the Moon’s Ann. Equ. Tab. VIII. + 0 22 Equ. Moon’s Anom.   - 20
And it gives the Mean time corrected 31 22 3 Anom. cor. 11 9 1
From which subtract the Moon’s elliptic Equ. Tab. X. - 3 10 Sun’s Equat. + 1 56
Moon’s Ano. 11 10 57
And it gives the Mean time equated 31 18 53   h. m.  
To which add the Sun’s Equation, Tab. XI. + 3 32 Moon’s ann. Equ. 0 22 add
    Her ellipt. Equ. 3 10 sub.
 
And it gives the true time of Conjunction 31 22 25 Sun’s Equation 3 32 add

Which true time answers to the first of April, at 25 minutes past 10 in the forenoon: for, as the Astronomical Day begins at Noon, then 22 hours 25 min. after the Noon of March 31, is April 1, at 10 hours 25 min. in the Forenoon.

EXAMPLE II.
 
To find the time of Full Moon in May 1761, N. S.

  Days Hours Min. Sun’s Anom. Moon’s Ano.
s ° ʹ s ° ʹ
Mean time of Full Moon in March 20 12 9 8 20 2 9 1 13
Add, for two Lunations 59 1 28 1 28 13 1 21 38
The several sums are 79 13 37 10 18 15 10 22 51
The days, in Tab. VII, answer to May 18 18 13 37 Equ. Moon’s Anom. - 13
Moon’s annual Equation add   + 14 Anom. cor. 10 22 38
Mean time corrected 18 13 51 Sun’s Equat. + 1 15
Moon’s elliptic Equation subtract - 5 38 Moon’s Ano. 10 23 53
Mean time equated 18 8 13   h. m.  
Sun’s Equation add + 2 19 Moon’s ann. Equ. 0 14 add
    Her ellipt. Equ. 5 38 sub.
 
True time of Opposition, May 18 10 32 Sun’s Equation 2 19 add

Namely, the 18th day, at 32 minutes past 10 at night.

The Leap-years are allowed for in the Tables, so as to give no Trouble in these Calculations.

To compute the time of New and Full Moon in a given year and month, of any particular Century, between the Christian Æra[78] and 18th Century.

Precept I. Find the like year of the 18th Century in Table I., for New Moon, or Table III., for Full Moon; and take out the New or Full Moon in March for that year, with the Anomalies of the Sun and Moon.

II. From Table V, take as many compleat Centuries, as when subtracted from the above year of the 18th Century, will answer to the given year; and take out the Conjunctions and Anomalies of these Centuries.

III. Subtract the Conjunctions and Anomalies of these Centuries from those of the New or Full Moon in March above taken out, and the remainders will shew the mean time of New or Full Moon in March the given year, with the Anomalies of the Sun and Moon at that time. Then, work in all respects for the true time of the proposed New or Full Moon, as taught by the Precepts already given § 355.

EXAMPLE I.
 
To find the time of New Moon in July 1581, O. S.

From 1781 subtract 200 years, and there remains 1581.

  Days Hours Min. Sun’s Anom. Moon’s Ano.
s ° ʹ s ° ʹ
Table I. Mean time of New Moon in March 1781 13 7 52 8 23 37 0 0 53
Tab. V. Conj. and Anom. for 200 years subtract 8 16 22 0 6 42 5 0 44
Remain the Conj. and Anom. for March 1581 4 15 30 8 16 55 7 0 9
Tab. VI. Add, for five Lunations, to bring it to July 147 15 40 4 25 32 4 9 5
The sums are 152 7 10 1 12 27 11 9 14
The Days in Tab. VII. answer to July 30th 30 7 10 Equ. Moon’s Anom. + 13
Sum of the three Equations subtract - 7 9 Anom. cor. 11 9 27
    Sun’s Equat. - 1 16
 
True time of Conjunction, July 30 0 1 Moon’s Ano. 11 8 11
Which is the 30th day, at one minute past noon, as shewn by well regulated Clocks or Watches Moon’s ann. Eq. 0h 14m sub.
Her ellipt. Equ. 3 35 sub.
Sun’s Equation 3 20 sub.
Sum 7 9 sub.
EXAMPLE II.
To find the time of Full Moon in April A. D. 30, O. S.
From 1730 subtract 1700, and there remains 30.
  Days Hours Min. Sun’s Anom. Moon’s Ano.
s ° ʹ s ° ʹ
Tab. III. Mean time of Full Moon in March 1730 22 6 58 9 2 40 3 13 23
Tab. V. Conj. and Anom. for 1700 years subtract 14 17 37 11 28 46 10 29 36
Rem. the Opposition and Anom. in March A. D. 30 7 13 21 9 3 54 4 13 47
Tab. V. Add, for one Lunation, to bring it into April 29 12 44 0 29 6 0 25 49
The sums are 37 2 5 10 3 0 5 9 36
The Days in Tab. VII. answer to April 6 6 2 5 Equ. Moon’s Anom. - 17
To which add the sum of the three Equations   6 1 Anom. cor. 5 9 19
    Sun’s Equat. + 1 35
 
True time of Opposition April A. D. 30 6 8 6 Moon’s Ano. 5 10 54
Which is the 6th day, at 6 minutes past 8 in the Evening. And thus, the time of New or Full Moon may be found for any given year and month after the Christian Æra.   Moon’s ann. Eq. 0h 18m add
Her ellipt. Equ. 2 46 add
Sun’s Equat. 2 57 add
Sum 6 1 add
Remark.

N. B. Sometimes it happens that the days annexed to the Centuries in Table V are more in number than the days on which the New or Full Moon happens in March the year of the 18th Century, with which the computation begins; as in the third following Example, viz. for the Full Moon in March the year before Christ 721: in which case, a Lunation and it’s Anomalies must be added, from Table VI, to the days and Anomalies of the New or Full Moon in March; and then, subtraction can be made: and having gained a remainder, work in all respects as taught in § 355.

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.
Age of the world uncertain.

By the method prescribed § 248 it will be found, that the Autumnal Equinox in the year before Christ 4030, fell on the 26th of October; as this Example shews the Full Moon to have been on the same day: and by working as hereafter taught, it will appear that the Dominical Letter was then G, which shews the 26th of that October to have been on a Friday; namely our sixth day of the week, but the Ante-Mosaic fifth day. And as, according to Genesis, chap. i. ver. 14. the Sun and Moon were created on the fourth day of the week, those who are of opinion that the world was made at the time of the Autumnal Equinox, and that the Moon at her first appearance was in full lustre, opposite to the Sun, or nearly so, may perhaps look upon this as a Criterion for ascertaining the year of the creation; since it shews the Moon to have been Full the next day after she was made: and this is only 9 years sooner than Rheinholt makes it, and 11 years later than according to Lange. Whereas, they who maintain that the world was created in the 4007th year before Christ, with the Sun on the Autumnal Equinoctial Point, October 26, and the Moon then Full; will find, if they compute by the best Tables extant, that the Moon was New, instead of being Full, on that day.

If it could be proved from the writings of Moses that the Sun was created on the point of the Autumnal Equinox, and the Moon in opposition; as well as it can be proved that these Luminaries were made (or according to some, did not shine out till) on the fourth day of the creation-week, there would be Data enough for ascertaining the age of the world: for supposing the Moon to have been Full on an Equinoctial Day, which was the fourth day of the week, it would require many thousands of years to bring these three characters together again. For, the soonest in which the Moon returns to be New or Full on the same days of the Months as before, is 19 years wanting an hour and half, but then the days of the week return not to the same days of the months in less than 28 years, in which time the Moon has gone through one Course of Lunations, and 9 years over; therefore a co-incidence of the Full Moon and day of the Week and Month cannot happen in that time, and if we multiply 19 by 28, which is the nearest co-incidence of these three characters, namely 532 years; the Moon’s falling back an hour and half every 19 years will amount to 42 hours in so many years; and the Equinox will have anticipated five days. From all which we may venture to say, that 200000 years would not be sufficient to bring all these circumstances together again.

EXAMPLE III.
 
To find the time of Full Moon at Babylon in March, the year before Christ 721.

The years 720 added to 1780 make 2500, or 25 Centuries.
  Days Hours Min. Sun’s Anom. Moon’s Ano.
s ° ʹ s ° ʹ
Tab. I. To the mean F. Moon and Anom. in Mar. 1780 9 4 41 8 19 48 7 8 10
Add one Lunation and it’s Anomalies from Tab. VI[80] 29 12 44 0 29 6 0 25 49
The several sums are 38 17 25 9 18 54 8 3 59
Fr. which subt. the Days & Anom. of 2500 years, Tab. V 19 22 20 11 26 19 6 6 43
Rem. the mean time and Anom. of F.M. in Mar. b.C. 721 18 19 5 9 22 25 1 27 16
To which add the sum of the three Equations + 11 36 Equ. Moon’s Anom. - 18
And it gives the true time of Full Moon, Mar. b.C. 721 18 6 41 Anom. cor. 1 26 48
    Sun’s Equat. + 1 47
Which was the 19th day, at 41 minutes past 6 in the evening, at London; to which time, if[81] 2 hours 51 minutes be added, we shall have the time at Babylon, namely, 9 hours 51 minutes.   Moon’s Anom. 1 28 35
Moon’s ann. Eq. 0h 20m add
Her ellipt. Equ. 8 1 add
Sun’s Equation 3 15 add
Sum 11 36 add

357. To know whether the Sun will be eclipsed or no, at the time of any given New Moon; collect the Sun’s distance from the Node at that time, and if it be less than 17 degrees he will be eclipsed, otherwise not.

EXAMPLE.
 
For the time of New Moon in April 1764.

  Sun from Node
s ° ʹ
Table II, mean New Moon in March 1764, New Stile, 11 4 57
Table VI, add for 1 Lunation to carry it to April 1 0 40
Sun’s distance from the Node at New Moon in April 0 5 37

Which, being within the above limit, the Sun must be eclipsed: and therefore, we proceed to find the rest of the Elements for computing this Eclipse.

To find the Moons Horizontal Parallax, or the Angle of the Earth’s semi-diameter as seen from the Moon.

Second Element.

358. Precept. Having found the Moon’s mean Anomaly for the above time, by the first and second Precepts of § 355, enter the XVth Table with the signs and degrees of that Anomaly, and thereby take out the Moon’s Horizontal Parallax: only note, that this is given but to every 6th degree of Anomaly in the Table, because it is very easy to make proper allowance by sight. So the Moon’s Horizontal Parallax April the 1st 1764, at 10 hours 25 minutes in the Forenoon, answering to her mean Anomaly at that time (namely 11s 9° 21ʹ) is 55ʹ 7ʺ; which, diminished by 10ʺ, the Sun’s constant Horizontal Parallax, gives for the semi-diameter of the Earth’s Disc 54ʹ 57ʺ.

To find the Sun’s true Place, and his distance from the nearest Solstice.

Third Element.

359. Precept I. We are to consider, that the beginning of Aries and of Libra, which are the Equinoctial Points, are equidistant from the beginning of Cancer and of Capricorn, which are the Solstitial Points. Hence, when we know in what Sign and Degree the Sun is, we can easily find his distance from the nearest Solstice. Now, to find the Sun’s Place, or Longitude from Aries, April the 1st, 1764, at 10 hours 21 minutes in the Forenoon; being the equated time of New Moon.

Precept II. This being to the time of New Moon, take out the Sun’s mean Place and Anomaly from Table II. for that time, and the Equation of his mean Place from Table XII by his Anomaly; adding the Equation to his mean Place or subtracting it from the same, as the Table directs, will give his true Place.

EXAMPLE.