Sun’s Long. from Aries. Sun’s mean 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.
s ° ʹ ʺ s ° ʹ
The year next less than 1757 in the Table is 1753, at the beginning of which, the 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
And likewise his mean Mot. and Anom. for   April 2 28 42 30 2 28 42
days 29 0 28 35 2 0 28 35
hours 22   0 54 13   0 54
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. ♉ Taurus 10° 11ʹ 26ʺ 1 10 11 26 1° 40ʹ 14ʺ

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
And also those of   May 3 28 16 40 3 28 17
days 28 Bissextile 0 28 35 2 0 28 35
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, viz. ♉ Taurus 29° 43ʹ 17ʺ 1 29 43 17 subtract.

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.

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 0s 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.

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.

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.

Plate XII.

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.

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.

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.

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.

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.

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.

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.

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.

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.

Axis of the Ecliptic.

371. Upon the center C raise the straight line CH for the Axis of the Ecliptic, perpendicular to ACB.

North Pole of the Earth.

372. Make a line of Chords to the Radius AC, and taking from thence the Chord of 231⁄2 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.

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 CXIIP, for the Earth’s Axis and the Universal Meridian; of both which P is the North Pole.

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) 511⁄2 degrees from 90 degrees, and there remains 381⁄2; 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 VILVI. 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 FXIIG parallel to VI L VI.

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 VIKV1 parallel to the occult or dotted Line VILVI. Then, making AC the Radius or length of a Line of Lines, set off the Sine of 381⁄2 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 7m, 8l, 9k, &c. all to the Line VIKVI, and parallel to CXIIP. Then, with KXII as a Radius, describe the Circle abcdef, round the Center K, and divide the Quadrant aXII 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 VIIbV, VIIIcIIII, IXdIII, &c. intersecting the former Lines 7m, 8l, 9k, 10i, &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 VIsVI 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 VIsVI would have been the diurnal path of London; in which case the Lines 7m, 8l, &c. must have been continued thro’ the right Line VIKVI, 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 VIIb, VIIIc &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.

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.

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.

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.

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.

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.

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.

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 10s 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
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
To project a lunar Eclipse.

Fig. III.

389. This done, make a Scale of any convenient length as W, whereof each division is a minute of a degree; and take from it in your Compasses 54 Minutes 50 Seconds, the Sum of Semi-diameters of the Moon and Earth’s shadow; and with that extent as a Radius, describe that Circle OVLG round C as a Center.

From the same Scale take 39 Minutes 48 Seconds, the Semi-diameter of the Earth’s shadow, and therewith as a Radius, describe the Circle UUUU for the Earth’s shadow, round C as a Center. Subtract the Moon’s Semi-diameter from the Semi-diameter of the Shadow, and with the difference 24 Minutes 46 seconds as a Radius, taken from the Scale W, describe the Circle YZ round the Center C.

Draw the right line AB through the Center C for the Ecliptic, and cross it at right Angles with the line EG for the Axis of the Ecliptic.

Because the Moon’s Latitude in this Eclipse is North Descending, § 384, set off the Angle of her visible Path with the Ecliptic 5 Degrees 38 Minutes (Page 202.) from E to V; and draw VCv for the Axis of the Moon’s Orbit. Had the Moon’s Latitude been North Ascending, this Angle must have been set off from E to f. N. B. When the Moon’s Latitude is South Ascending, the Axis of her Orbit lies the same way as when she has North Ascending Latitude; and when her Latitude is North Descending, the Axis of her Orbit lies the same way as when her Latitude is South Descending.

Take the Moon’s true Latitude 9ʹ 56ʺ in your Compasses from the Scale W, and set it off from C to F on the Axis of the Ecliptic because the Moon is north of the Ecliptic; (had she been to the South of it, her Latitude must have been set off the contrary way, as from C towards v:) and through F, at right Angles to the Axis of the Moon’s Orbit VCv, draw the right line LMHNO for the Moon’s Orbit, or her Path through the Earth’s shadow. N. B. When the Moon’s Latitude is North Ascending, or North Descending, she is above the Ecliptic: but when her Latitude is South Ascending, or South Descending, she is below it.

Take the Moon’s true horary motion from the Sun, viz. 28 Minutes 22 Seconds, from the Scale W in your Compasses; and with that extent make marks in the line of the Moon’s Path LMHNO: then divide each of these equal spaces into 60 equal parts or minutes of time: and set the hours to them as in the Figure, in such a manner that the precise time of Full Moon, as shewn by the Tables, may fall in the Axis of the Ecliptic at F, where the line of the Moon Path cuts it.

Lastly, Take the Moon’s Semi-diameter 15 Minutes 2 Seconds from the Scale W in your Compasses, and therewith as a Radius describe the Circles P, Q, R, S, and T on the Centers L, M, H, N, and O; the Circles P and T just touching the Earth’s Shadow UU, but no part of them within it; the Circles Q and S all within it, but touching at its edges; and the Circle R in the middle of the Moon’s Path through the shadow. So the Circle P shall be the Moon touching the shadow at the moment the Eclipse begins; the Circle Q the Moon just immersed into the shadow at the moment she is totally eclipsed; the Circle R the Moon at the greatest obscuration, in the middle of the Eclipse; the Circle S the Moon just beginning to be enlightened on her western limb at the end of total darkness; and the Circle T the Moon quite clear of the Earth’s shadow at the moment the Eclipse ends. The moments of time marked at the points L, M, H, N and O answer to these Phenomena: and according to this small projection are as follow. The beginning of the Eclipse at 8 Hours 36 Minutes P. M. The total immersion at 9 Hours 42 Minutes. The middle of the Eclipse at 10 Hours 26 Minutes. The end of total darkness at 11 Hours 12 Minutes. And the end of the Eclipse at 12 Hours 18 Minutes; but the Figure is too small to admit of precision.