Chapter XV: Part 15

+------------------------------------+----------+-------------+-------------+ | | | Sun’s Anom. | Moon’s Ano. | | | D. H. M. +-------------+-------------+ | | | 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 | | Sun’s Equat. + 1 56 | | elliptic Equ. Tab. X. | - 3 10 | ---------- | | | -------- | Moon’s Ano. 11 10 57 | | And it gives the Mean time equated | 31 18 53 +---------------------------+ | To which add the Sun’s Equation, | | h. m. | | Tab. XI. | + 3 32 | Moon’s ann. Equ. 0 22 add | | And it gives the true time | -------- | Her ellipt. Equ. 3 10 sub.| | 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._

+------------------------------------+----------+-------------+-------------+ | | | Sun’s Anom. | Moon’s Ano. | | | D. H. M. +-------------+-------------+ | | | 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 | | Equ. Moon’s Anom. - 13 | | _May 18_ | 18 13 37 | ---------- | | Moon’s annual Equation add | + 14 | Anom. cor. 10 22 38 | | | -------- | Sun’s Equat. + 1 15 | | Mean time corrected | 18 13 51 | ---------- | | 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.

+-----------------------------------+-----------+-------------+-------------+ | | | Sun’s Anom. | Moon’s Ano. | | | D. H. M. +-------------+-------------+ | | | 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 | | Equ. Moon’s Anom. + 13 | | to _July_ 30th | 30 7 10 | ----------- | | Sum of the three Equations | | Anom. cor. 11 9 27 | | subtract | - 7 9 | 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 | Moon’s ann. Eq. 0^h 14^m sub. | | past noon, as shewn by well | Her ellipt. Equ. 3 35 sub. | | regulated Clocks or Watches | 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.

+------------------------------------+----------+-------------+-------------+ | | | Sun’s Anom. | Moon’s Ano. | | | D. H. M. +-------------+-------------+ | | | 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 | | Equ. Moon’s Anom. - 17 | | _April 6_ | 6 2 5 | ---------- | | To which add the sum of the three | | Anom. cor. 5 9 19 | | Equations | 6 1 | Sun’s Equat. + 1 35 | | | -------- | ---------- | | True time of Opposition | | Moon’s Ano. 5 10 54 | | _April_ A. D. 30 | 6 8 6 | | | +-----+----+---------------------------+ | Which is the 6th day, at 6 minutes past | Moon’s ann. Eq. 0^h 18^m add | | 8 in the Evening. And thus, the time | Her ellipt. Equ. 2 46 add | | of New or Full Moon may be found for | Sun’s Equat. 2 57 add | | any given year and month after the | ------------- | | Christian Æra. | Sum 6 1 add | +------------------------------------------+--------------------------------+

[Sidenote: 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.

+------------------------------------+----------+-------------+-------------+ | | | Sun’s Anom. | Moon’s Ano. | | | D. H. M. +-------------+-------------+ | | | 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 | | | -------- | ---------- | | Gives the true time of Conjunction | 20 19 26 | Anom. cor. 1 4 55 | | +----------+ Sun’s Equat. + 1 48 | | Which was the 21st day of _March_, at | --------- | | 26 minutes past 7 in the morning at | Moon’s Ano. 1 6 43 | | _London_: and if 1 hour 35 minutes +---+---------------------------+ | be added for _Athens_, which is 23° 52ʹ | Moon’s ann. Eq. 0^h 20^m add | | east of the meridian of _London_, we | Her ellipt. Equ. 5 43 add | | have the time at _Athens_; namely, | Sun’s Equation 3 17 add | | 1 minute past 9 in the morning. | 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.

+-----------------------------------+-----------+-------------+-------------+ | | | Sun’s Anom. | Moon’s Ano. | | | D. H. M. +-------------+-------------+ | | | 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. | | Rem. the true time of Full Moon, | | Eq. 3 28 sub. | | _October_ | 25 20 56 | Sun’s Equation 0 0 add | | +-----------+ --------- | | Which is the 26th day, at 8 hours | Total 3 28 sub. | | 26 minutes in the forenoon[79]. | | +-----------------------------------------------+---------------------------+

[Sidenote: 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.

+------------------------------------+----------+-------------+-------------+ | | | Sun’s Anom. | Moon’s Ano. | | | D. H. M. +-------------+-------------+ | | | 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 | | | -------- | Anom. cor. 1 26 48 | | And it gives the true time of | | Sun’s Equat. + 1 47 | | Full Moon, _Mar. b.C. 721_ | 18 6 41 | ---------- | | +------+---+ Moon’s Anom. 1 28 35 | | Which was the 19th day, at 41 minutes +-------------------------------+ | past 6 in the evening, at _London_; | Moon’s ann. Eq. 0^h 20^m add | | to which time, if[81] 2 hours 51 | Her ellipt. Equ. 8 1 add | | minutes be added, we shall have | Sun’s Equation 3 15 add | | the time at _Babylon_, namely, | Sum 11 36 add | | 9 hours 51 minutes. | | +-------------------------------------------+-------------------------------+

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._

[Sidenote: 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 11^s 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._

[Sidenote: 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.

+----------------------------------------------+-------------+------------+ | | Sun’s Long. | Sun’s mean | | | from Aries. | Anomaly. | | +-------------+------------+ | | s ° ʹ | s ° ʹ | | Table I. To the Sun’s mean Place and +-------------+------------+ | Anomaly at the mean time of New Moon | | | | in _March 1764_, N. S. | 11 17 7 | 8 2 23 | | Add the same from Tab. VI. for one Lunation, | | | | to carry it to _April_ | 0 29 6 | 0 29 6 | | | --------- | ---------- | | Mean Place and Anomaly at the time of New | | | | Moon in _April_ | 0 10 13 | 9 1 29 | | To which place add the Sun’s Equation | +------------+ | from Tab. XII. | 1 56 | Equal | | | --------- | 1° 56ʹ | | And it gives the Sun’s true place | 0 12 9 | Additive. | | +-------------+------------+ | Which is Aries 12° 9ʹ; and this, when taken from three Signs, or the | | beginning of Cancer, leaves 2 signs 17 deg. 51 min., or 77° 51ʹ for | | the Sun’s distance from the then nearest Solstice. | +-------------------------------------------------------------------------+

360. But because the Sun’s true Place is often wanted when the Moon is neither New nor Full, we shall next shew how it may be found for any given moment of time: though this be digressing from our present purpose.

In Table XVI find the nearest lesser year to that in which the Sun’s Place is sought; and take out the Sun’s mean Longitude and Anomaly answering thereto; to which add his mean motion and Anomaly for the compleat residue of the years, with the month, day, hour, and minute, all taken from the same Table, and you have the Sun’s mean Longitude and Anomaly for the given time. Then, from Table XII take out the Sun’s Equation by means of his Anomaly (making proportions for the odd minutes of Anomaly) which Equation being added to or subtracted from the Sun’s mean Longitude from Aries, as the titles in the Table direct, gives his true Place, or Longitude from the beginning of Aries, reckoned according to the order of the Signs § 354.

EXAMPLE.

_To find the Sun’s true Place_ April _30th, A. D. 1757, at 18 minutes 40
seconds past 10 in the morning_.

+---------------------------------------------+-------------+-------------+ | | Sun’s Long. | Sun’s Anom. | | The year next less than 1757 in the Table +-------------+-------------+ | is 1753, at the beginning of which, the | s ° ʹ ʺ | s ° ʹ | | Sun’s mean Longitude from the beginning +-------------+-------------+ | of Aries, and his mean Anomaly, is | 9 10 16 52 | 6 1 38 | | To which add his mean Mot. and Anom. for | | | | four years to make 1757 | 0 0 1 49 | 11 29 58 | | { _April_ | 2 28 42 30 | 2 28 42 | | { days 29 | 0 28 35 2 | 0 28 35 | | And likewise his mean Mot. and { hours 22 | 0 54 13 | 0 54 | | Anom. for { min. 18 | 0 44 | 1 | | { sec. 49 | 2 | 0 | | | ----------- |-------------+ | Sun’s mean Longitude and Anomaly for the | | | | given time is | 1 8 31 12 | 9 29 48 | | To which add the Equation of the Sun’s | | | | mean Place | 1 40 14 +-------------+ | | ----------- | Sun’s Eq. | | And it gives his true Place, _viz._ | | 1° 40ʹ 14ʺ | | ♉ Taurus 10° 11ʹ 26ʺ | 1 10 11 26 | | +---------------------------------------------+-------------+-------------+

N. B. _In leap-years after_ February, _the Sun’s mean Motion and Anomaly must be taken out for the day next after the given one._

361. _To calculate the Sun’s true Place for any time in a given year before the first year of_ CHRIST: subtract the mean Motions and Anomalies for the compleat hundreds next above the given year; to the remainder add those for the residue of years, months, _&c._ and then work in all respects as above taught.

EXAMPLE.

_To find the Suns true Place_ May _the 28th at 4 hours 3 min. 42 sec. in the afternoon, the year before Christ 585, which was a Leap year_[82].

+---------------------------------------------+-------------+-------------+ | | Sun’s Long. | Sun’s Anom. | | +-------------+-------------+ | | s ° ʹ ʺ | s ° ʹ | | From the Sun’s mean Longitude and Anomaly +-------------+-------------+ | at the beginning of the year Christ 1 | 9 7 53 10 | 6 29 54 | | Subtract his mean Motion and Anomaly for | | | | 600 years | 0 4 32 0 | 11 24 2 | | + ----------- | ---------- | | And the remainder, or radix, is | 9 3 21 10 | 7 5 52 | | To which add what 585 wants of 600, | | | | _viz._ 15 years | 11 29 22 27 | 11 29 7 | | { _May_ | 3 28 16 40 | 3 28 17 | | { days 28 Bissextile | 0 28 35 2 | 0 28 35 | | And also those of { hours 4 | 0 9 51 | 0 10 | | { min. 3 | 0 7 | ---------- | | { sec. 42 | 2 | 0 2 1 | | | ----------- | Sun’s Anom. | | Sun’s mean Long. _May_ 28th, at 4 hour | +-------------+ | 3 min. 24 sec. afternoon | 1 29 45 19 | | | Equation of the Sun’s mean Place subtract | 2 2 | 2ʹ 22ʺ | | | ----------- | Sun’s Equat.| | Rem. his true Place for the same time, | | subtract. | | _viz._ ♉ Taurus 29° 43ʹ 17ʺ | 1 29 43 17 | | +---------------------------------------------+-------------+-------------+

_N. B._ As the Longitudes or Places of all the visible Stars in the Heavens are well known, we have an easy method of finding the Sun’s true Place in the Ecliptic: for the Sun is directly opposite to that Point of the Ecliptic which comes to the Meridian at mid-night.

_To find the Sun’s Declination._

[Sidenote: Fourth Element.]

362. PRECEPT. Enter Table XVII with the Signs and Degrees of the Sun’s Place; and making proportions, take out his Declination answering thereto. If the Signs are at the head of the Table, the Degrees are at the left hand; but if the Signs are at the foot of the Table, the Degrees are at the right hand. So, the Sun’s Declination answering to his true Place (found by § 359 to be 0^s 12° 9ʹ) is 4 degrees 48 minutes 54 seconds, making allowance for the 9ʹ that his Place exceeds 12°.

_To find the Angle of the Moon’s visible Path with the Ecliptic._

[Sidenote: Fifth Element.]

PRECEPT. This we may state at 5 degrees 38 minutes, as near enough for the purpose; since it is never above 8 minutes of a degree more or less.

_To find the Moon’s Latitude._

[Sidenote: Sixth Element.]

363. PRECEPT. Having found the Sun’s distance from the Ascending Node by § 357, at the mean time of New Moon, and his Anomaly for that time by § 359, find the Equation of the Node in Table XIII, by the Sun’s Anomaly, and the Equation of the Sun’s mean Place in Table XII by his Anomaly: these two Equations applied (as the titles direct) to the Sun’s mean distance from the Ascending Node, give his true distance from it, and also the Moon’s true distance at the time of Change: but when the Moon is Full, this distance must be increased by the addition of 6 Signs, which will then be the Moon’s true distance from the same Node.

The Moon’s true distance from the Ascending Node is called the _Argument of the Moon’s Latitude_; with the Signs of which, at the head of Table XIV, and Degrees at the left hand, or with the Signs at the foot of the Table and Degrees at the right hand, take out the Moon’s Latitude: which is _North Ascending_, _North Descending_, _South Ascending_, or _South Descending_, according to the letters _NA_, _ND_, _SA_ or _SD_, annexed to the Signs of the said Argument.

_The Geometrical Construction of Solar and Lunar Eclipses._

_J. Ferguson delin._ _J. Mynde Sculp._]

EXAMPLE.

s ° ʹ Sun’s mean Dist. from the [83]Node at New Moon in _April 1764_ 0 5 37 To which add the Equation of the Node + 10 ---------- And it gives the Sun’s corrected Distance from the Node 0 5 47 To which cor. Dist. add the Eq. of the Sun’s mean Place + 1 56 ---------- And it gives the Sun’s true Distance from the Node 0 7 43

Which, being at the time of New Moon, is the _Argument of Latitude_; and in Table XIV, (making proportions for the 43ʹ) shews the Moon’s Latitude to be 40ʹ 9ʺ _North Ascending_[84].

_To find the Moon’s true hourly Motion from the Sun._

[Sidenote: Seventh Element.]

364. PRECEPT. With the Moon’s Anomaly enter Table XV, and thereby take out her true hourly Motion: then with the Sun’s Anomaly take out his true hourly Motion from the same Table: which done, subtract the Sun’s hourly Motion from the Moon’s, and the remainder will be the Moon’s true hourly Motion from the Sun; which, for the above time § 359, is 27ʹ 50ʺ.

_To find the Semi-diameters of the Sun and Moon as seen from the Earth
at the above-mentioned time._

[Sidenote: Eighth and Ninth Elements.]

365. PRECEPT. Enter the XVth Table with the Sun’s Anomaly, and thereby take out his Semi-diameter; and in the same manner take out the Moon’s Semi-diameter by her Anomaly. The former of which for the above time will be found to be 16ʹ 6ʺ; the latter 14ʹ 58ʺ.

_To find the Semi-diameter of the Penumbra._

[Sidenote: Tenth Element.]

366. PRECEPT. Add the Sun’s semi-diameter to the Moon’s, and their Sum will be the Semi-diameter of the Penumbra; namely, at the above time 31ʹ 4ʺ.

[Sidenote: Pl. XII.]

366. Having found the proper Elements or Requisites for the Sun’s Eclipse _April 1, 1764_, and intending to project this Eclipse Geometrically, we shall now collect them under the eye, that they may be the more readily found as they are wanted in order for the Projection.

[Sidenote: The proper Elements collected.]

D H M

367. I. The true time of Conj. or New Moon _April_ 1 10 25

° ʹ ʺ

II. The Earth’s Semi-Disc, which is equal to the
Moon’s Horizontal Parallax 55ʹ 7ʺ diminished by
the Sun’s Horizontal Parallax which is always 10ʺ 0 54 57

III. The Sun’s distance from the nearest Solstice,
_viz._ ♋ 77 51 0

IV. The Sun’s Declination, North 4 48 54

V. The Angle of the Moon’s vis. path with the
Eclipt. 5 38 0

VI. The Moon’s true Latitude, North Ascending 40 9

VII. The Moon’s true Horary Motion from the Sun 27 50

VIII. The Sun’s Semi-diameter 16 6

IX. The Moon’s Semi-diameter 14 58

X. The Semi-diameter of the Penumbra 31 4

368. Having collected these Elements or Requisites, the following part of the work may be very much facilitated by means of a good Sector, with the use of which the reader should be so well acquainted, as to know how to open it to any given Radius, as far as it will go; and to take off the Chord or Sine of any Arc of that Radius. This is done by first taking the extent of the given Radius in your Compasses, and then opening the Sector so as the distance cross-wise between the ends of the lines of Sines or Chords at _S_ or _C_, from Leg to Leg of the Sector, may be equal to that extent; then, without altering the Sector, take the Sine or Chord of the given Arc with your Compasses extended cross-wise from Leg to Leg of the Sector in these lines. But if the operator has not a Sector, he must construct these lines to such different lengths as he wants them in the projection. And lest this Treatise should fall into the hands of any person who would wish to project the Figure of a solar or lunar Eclipse, and has not a Sector to do it by, we shall shew how he may make a line of Sines or Chords to any Radius.

[Sidenote: Fig. II.

How to make a line of Chords.

Pl. XII.]

369. Draw the right line _BCA_ at pleasure; and upon _C_ as a Center, with the distance _CA_ or _CB_ as a Radius, describe the Semi-circle _BDA_; and from the Center _C_ draw _AC_ perpendicular to _BCA_. Then divide the Quadrants _AD_ and _BD_ each into 90 equal parts or degrees, and join the right line _AD_ for the Chord of the Quadrant _AD_. This done, setting one foot of the Compasses in _A_, extend the other to the different divisions of the Quadrant _AD_; and so transfer them to the right line _AD_ as in the Figure, and you have a line of Chords _AD_ to the Radius _CA_. _N. B._ 60 Degrees on the Line of Chords is always equal to the Radius of the Circle it is made from; as is evident by the Figure, where the Arch _E_, whose Center is _A_, drawn from 60 on the Quadrant _AD_, cuts the Chord line in 60 degrees, and terminates in the Center _C_.

[Sidenote: And of Sines.]

Then, from the divisions or degrees of the Quadrant _BD_, draw lines parallel to _CD_, which will fall perpendicularly on the Radius _BC_, dividing it into a line of Sines; and it will be near enough for the present purpose, to have them to every fifth Degree, as in the Figure. And thus the young _Tyro_ may supply himself with Chords and Sines, if he has not a Sector. But as the Sector greatly shortens the work, we shall describe the projection as done by it, so far as Signs and Chords are required.

[Sidenote: Fig. II.

Earth’s Semi-Disc.]

370. Make a Scale of any convenient length (six inches at least) as _AC_, and divide it into as many equal parts as the semi-diameter of the Earth’s Disc contains minutes, which in this construction of the Eclipse for _London_ in _April 1764_, is 54 minutes and 57 seconds; but as it wants only 3ʺ of 55ʹ the Scale may be divided into 55 equal parts, as in the Figure. Then, with the whole length of the Scale as a Radius, setting one foot of your Compasses in _C_ as a center, describe the Semi-circle _AMB_ for the northern Hemisphere or Semi-disc of the Earth, as seen from the Sun at that time. Had the Place for which the Construction is made been in South Latitude, this Semi-circle would have been the Southern Hemisphere of the Earth’s Disc.

[Sidenote: Axis of the Ecliptic.]

371. Upon the center _C_ raise the straight line _CH_ for the Axis of the Ecliptic, perpendicular to _ACB_.

[Sidenote: North Pole of the Earth.]

372. Make a line of Chords to the Radius _AC_, and taking from thence the Chord of 23-1/2 Degrees, set it off from _H_ to _g_ and to _h_, on the periphery of the Semi-disc; and draw the straight line _gNh_, in which the North Pole of the Disc is always found.

373. While the Sun is in Aries, Taurus, Gemini, Cancer, Leo, and Virgo, the North Pole of the Disc is illuminated; but while the Sun is in Libra, Scorpio, Sagittary, Capricorn, and Aquarius, the North Pole is hid in the obscure part behind the Disc.

374. And, whilst the Sun is in Capricorn, Aquarius, Pisces, Aries, Taurus, and Gemini, the Earth’s Axis _CP_ lies to the right hand of the Axis of the Ecliptic _CH_ as seen from the Sun, and to the left hand while the Sun is in the other six Signs.

[Sidenote: Earth’s Axis.

Universal Meridian.]

375. Make a line of Sines equal in length to _Ng_ or _Nh_, and take off with your Compasses from it the Sine of the Sun’s distance from the nearest Solstice, which in the present case is 77° 51ʹ § 367, and set that distance to the right hand, from _N_ to _P_, on the line _gNh_, because the Sun being in Aries § 359, the Earth’s Axis lies to the right hand of the Axis of the Ecliptic § 374: then draw the straight line _C_XII_P_, for the Earth’s Axis and the Universal Meridian; of both which _P_ is the North Pole.

[Sidenote: Path of a given Place on the Disc as seen from the Sun.]

376. To draw the parallel of Latitude of any given Place (suppose _London_) which parallel is the visible Path of the Place On the Disc, as seen from the Sun, from the time that the Sun rises till it sets; subtract the Latitude of the Place (_London_) 51-1/2 degrees from 90 degrees, and there remains 38-1/2; which take from the Line of Chords in your Compasses, and set it from _h_ (where the Universal Meridian _CP_ cuts the periphery of the Semi-disc) to VI and VI; and draw the occult Line VI_L_VI. Then, on the left hand of the Earth’s Axis, set off the Chord of the Sun’s Declination 4° 48ʹ 5ʺ § 367, from VI to _D_ and to _F_; set off the same on the right hand from VI to _E_ and to _G_; and draw the occult Lines _DsE_ and _F_XII_G_ parallel to VI _L_ VI.

[Sidenote: Situation of the Place on the Disk from Sun-rise to Sun-set.]

377. Bisect _s_ XII in _K_, and through the point _K_ draw the black Line VI_K_V1 parallel to the occult or dotted Line VI_L_VI. Then, making _AC_ the Radius or length of a Line of Lines, set off the Sine of 38-1/2 degrees, the Co-Latitude of _London_, from _K_ to VI and VI; and with that extent as a Radius, describe the Semi-Circle VI 7 8 9 &c. and divide it into 12 equal parts, beginning at VI. From these divisions, draw the occult Lines 7_m_, 8_l_, 9_k_, &c. all to the Line VI_K_VI, and parallel to _C_XII_P_. Then, with _K_XII as a Radius, describe the Circle _abcdef_, round the Center _K_, and divide the Quadrant _a_XII into six equal parts, as _ab_, _bc_, _cd_, _de_, &c. Then, through these points of division _b_, _c_, _d_, _e_, and _f_, draw the occult Lines VII_b_V, VIII_c_IIII, IX_d_III, &c. intersecting the former Lines 7_m_, 8_l_, 9_k_, 10_i_, &c. in the points VII, VIII, IX, X, XI, &c. which points mark the situation of _London_ on the Earth’s Disc as seen from the Sun at these hours respectively, from six in the morning till six at night: and if the elliptic Curve VI, VII, VIII, &c. be drawn through these points, it will represent the parallel of _London_, or the path it seems to describe as viewed from the Sun, from Sun-rise to Sun-set. _N.B._ When the Sun’s Declination is North, the said Curve is the diurnal Path of _London_; and the opposite part VI_s_VI is it’s nocturnal Path behind the Disc, or in the obscure part thereof, § 338, 339. But if the Sun’s Declination had been South, the Curve VI_s_VI would have been the diurnal path of _London_; in which case the Lines 7_m_, 8_l_, &c. must have been continued thro’ the right Line VI_K_VI, and their lengths beyond that line determined by dividing the Quadrant _s a_ of the little Circle _abcd_ into six equal parts, and drawing the parallels VII_b_, VIII_c_ &c. through that division, in the same manner as done on the side _K_ XII; and the Curve VII, VIII, IX, &c. would have been the nocturnal Path. It is requisite to divide the hours of the diurnal Path into quarters, as in the Diagram; and if possible into minutes also.

[Sidenote: Axis of the Moon’s Orbit.]

378. From the Line of Chords § 372 take the Angle of the Moon’s visible Path with the Ecliptic, _viz._ 5° 38ʹ § 367: and note, that when the Moon’s Latitude is _North Ascending_, as in the present case, the Chord of this Angle must be set off to the left hand of the Axis of the Ecliptic _CH_, as from _H_ to _M_, and the right line _CM_ drawn for the Axis of the Moon’s Orbit: but when the Moon’s Latitude is _North Descending_, this Angle and Axis must be set to the right hand, or from _H_ toward _h_. When the Moon’s Latitude _South Ascending_, the Axis of her Orbit lies the same way as when her Latitude is _North Ascending_; and when _South Descending_, the same way as when _North Descending_.

[Sidenote: Path of the Penumbra’s center over the Earth.]

379. Take the Moon’s Latitude, 40ʹ 9ʺ § 367, from the Scale _CA_, and set it from _C_ to _T_ on the Axis of the Ecliptic; and through _T_, at right Angles to the Axis of the Moon’s Orbit _CM_, draw the straight Line _RTS_; which is the Moon’s Path, or Line that the center of her shadow and Penumbra describes in going over the Earth’s Disc. The Point _T_ in the Axis of the Ecliptic is the Place where the true Conjunction of the Sun and Moon falls, according to the Tables; and the Point _W_, in the Axis of the Moon’s Orbit, is that where the center of the Penumbra approaches nearest to the center of the Earth’s Disc, and consequently the middle of the general Eclipse.

[Sidenote: It’s Place on the Earth’s Disc shewn for every minute of it’s
Transit.]

380. Take the Moon’s true Horary Motion from the Sun 27ʹ 50ʺ § 367, from the Scale _CA_ with your Compasses (every division of the Scale being a minute of a Degree) and with that extent make marks in the Line of the Moon’s Path _RTS_: then divide each of these equal spaces by dots into 60 equal parts or horary minutes, and set the hours to every 60th minute, in such a manner that the dot; signifying the precise minute of New Moon by the Tables, may fall in the Point _T_ where the Axis of the Ecliptic cuts the Line of the Moon’s Path; which, in this Eclipse, is the 25th minute past ten in the Forenoon: and then the other marks will shew the places on the Earth’s Disc where the center of the Penumbra is, at the hours and minutes denoted by them, during its transit over the Earth.

[Sidenote: Middle of the Eclipse.

It’s Phases.]

381. Apply one side of a Square to the Line of the Moon’s Path, and move the Square backward or forward until the other side cuts the same hour and minute both in the Path of the Place (_London_, in this Construction) and Path of the Moon; and _that_ minute, cut at the same time in both Paths, will be the precise minute of visible Conjunction of the Sun and Moon at _London_, and therefore the time of greatest obscuration, or middle of the Eclipse at _London_; which time, in this Projection, falls at _t_, 34 minutes past 10 in the Moon’s Path; and at _u_, 34 minutes past 10 in the Path of _London_. Then, upon the Point _u_ as a center, describe the Circle _zYy_ whose Radius _uy_ is equal to the Sun’s semi-diameter 16ʹ 6ʺ § 367, taken from the Scale _CA_: And upon the Point _t_ as a center, describe the Circle _Hy_ whose Radius is equal to the Moon’s semi-diameter 14ʹ 58ʺ § 367, taken from the same Scale. The Circle _zYy_ represents the Disc of the Sun as seen from the Earth, and the Circle _Hy_ the Disc of the Moon. The portion of the Sun’s Disc cut off by the Moon’s shews the Quantity of the Eclipse at the time of greatest obscuration: and if a right Line as _yz_ be drawn across the Sun’s Disc through _t_ and _u_, the minute of greatest obscuration in both Paths, and divided into 12 equal parts, it will shew what number of Digits are then eclipsed. If these two Circles do not touch one another, the Eclipse will not be visible at the given Place.

[Sidenote: It’s beginning and ending.]

382. Lastly, take the Semi-diameter of the Penumbra 31ʹ 4ʺ § 367, from the Scale _CA_ with your Compasses; and setting one foot in the Moon’s Path, to the left hand of the Axis of the Ecliptic, direct the other toward the Path of _London_; and carry this extent backwards or forwards until both Points of the Compasses fall into the same instants of time in both Paths: which will denote the time of the beginning of the Eclipse: then, do the same on the right hand of the Axis of the Ecliptic, and where both Points mark the same instants in both Paths, they will shew at what time the Eclipse ends. These trials give the Points _R_ in the Moon’s Path and _r_ in the Path of _London_, namely 9 minutes past 9 in the Morning for the beginning of the Eclipse at _London_, _April 1, 1764_: _t_ and _u_ for the middle or greatest obscuration, at 35 minutes past 10; when the Eclipse will be barely annular on the Sun’s lower-most edge, and only two thirds of a Digit left free on his upper-most edge: and for the end of the Eclipse, _S_ in the Moon’s Path and _x_ in the Path of _London_, at 4 minutes past 12 at Noon.

In this Construction it is supposed that the Equator, Tropics, Parallel of _London_, and Meridians through every 15th degree of Longitude are projected in visible Lines on the Earth’s Disc, as seen from the Sun at almost an infinite distance; that the Angle under which the Moon’s diameter is seen, during the time of the Eclipse, continues invariably the same; that the Moon’s motion is uniform, and her Path rectilineal, for that time. But all these suppositions do not exactly agree with the truth; and therefore, supposing the Elements § 367, given by the Tables to be perfectly accurate, yet the time and phases of the Eclipse deduced from it’s Construction will not answer exactly to what passeth in the Heavens; but may be two or three minutes wrong though done with the utmost care. Moreover, the Paths of all Places of considerable Latitude go nearer the center of the Disc as seen from the Moon than these Constructions make them; because the Earth’s Disc is projected as if the Earth were a perfect sphere, although it is known to be a spheroid. Consequently, the Moon’s shadow will go farther North in places of northern Latitude, and farther South in places of southern Latitude than these projections answer to. Hence we may venture to predict that this Eclipse will be more annular at _London_ (that is, the annulus will be somewhat broader on the southern Limb of the Sun) than the Diagram shews it.

383. Having shewn how to compute the times and project the phases of a Solar Eclipse, we now proceed to those of the Lunar. And it has been already mentioned § 317, that when the Full Moon is within 12 degrees of either of her Nodes, she must be eclipsed. We shall now enquire whether or no the Moon will be eclipsed _May 18, 1761, N. S._ at 32 minutes past 10 at Night. See page 193.

[Sidenote: Table IV.

Table VI.]

s ° ʹ Sun from Node at Full Moon in _March 1761_ 9 25 27 Add his distance for two Lunations, to bring it into _May_ 2 1 20 --------- And his distance at Full Moon in that month is 11 26 47

Subtract this from a Circle, or 12 Signs, and there will remain 3° 13ʹ; which is all that the Sun wants of coming round to the Ascending Node; and the Moon being then opposite to the Sun, must be just as near the Descending Node: consequently, far within the limit of an Eclipse.

384. Knowing then that the Moon will be eclipsed in _May 1761_, we must find her true distance from the Node at that time, by applying the proper Equations as taught § 363, and then find her true Latitude as taught in that article.

[Sidenote: Table IV.

Table XIII.

Table XII.]

s ° ʹ Sun’s mean distance from the Node at F. Moon in _May 1761_ 11 26 47 Add the Equation of the Node, for the Sun’s Anomaly 10^s 18° 15ʹ[85] + 6 -------- Sun’s mean distance from the Node corrected 11 26 53 Add the Equation of the Sun’s mean Place + 1 15 -------- Sun’s true distance from the Ascending Node 11 28 8 To which add 6 Signs, See § 363 6 -------- The sum is the Moon’s true distance from the same Node 5 28 8

[Sidenote: Pl. XII.]

Or the _Argument_ of her _Latitude_; which in Table XIV, gives the Moon’s true Latitude, _viz._ 9ʹ 56ʺ North Descending.

385. Having by the foregoing precepts § 355 found the true time of Opposition of the Sun and Moon in a lunar Eclipse, with the Moon’s Anomaly enter Table XV and take out her horizontal Parallax, also her true horary Motion and Semi-diameter: and likewise those of the Sun by his Anomaly, as already taught § 364 & _seq._ Then add the Sun’s horizontal Parallax, which is always 10 Seconds, to the Moon’s horizontal Parallax, and from their sum subtract the Sun’s Semi-diameter; the remainder will be the Semi-diameter of that part of the Earth’s shadow which the Moon goes through.

386. From the Sum of the Semi-diameters of the Moon and Earth’s Shadow, subtract the Moon’s Latitude; the remainder is the parts deficient. Then, as the Semi-diameter of the Moon is to 6 Digits, so are the parts deficient to the Digits eclipsed.

387. If the parts deficient be more than the Moon’s Diameter, the Eclipse will be total with continuance; if less, it will not be total; if equal, it will be total, but without continuance.

388. Now collect the Elements for projecting this Eclipse.

ʹ ʺ
Moon’s horizontal Parallax 55 32
Sun’s horizontal Parallax (always) 10
The Sum of both Parallaxes 55 42
From which subtract the Sun’s Semi-diameter 15 54
Remains the Semi-diameter of the Earth’s Shadow 39 48
Semidiameter of the Moon 15 2
Sum of the two last 54 50
Moon’s Latitude subtract 9 56
Remains the parts deficient 45 0
Moon’s horary motion 30 46
Sun’s horary motion subtract 2 24
Remains the Moon’s horary motion from the Sun 28 22

[Sidenote: To project a lunar Eclipse.

Fig. III.]