Part 29

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.

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.]