Part 28
+------------------------------------+----------+-------------+-------------+ | | | 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.]