Mathematical Instruments
Book VIII.

# The Use of the Sector in the Construction of Solar Eclipses.

## Definition I.

The Path of a Vertex, is that Circle of the Earth which any Place or Vertex on it’s Superficies describes, in the Space of twenty-four Hours, by the Earth’s diurnal Revolution. Whence the Paths of Verticies are Circles parallel to the Equator.

## Definition II.

If a Plane be conceived to touch the Moon’s Orbit in that Point, wherein a Line connecting the Centers of the Earth and Sun intersects the said Orbit, and Hands at right Angles to the aforenamed Line And if an infinite Number of right Lines be supposed to pass from the Center of the Sun, thro’ this Plane to the Periphery of the Earth, to it’s Axis, as likewise to the Axis of the Ecliptick, and the Path of any Vertex; the said Lines will orthographically project the Earth’s Disk, it’s Axis, the Axis of the Ecliptick, and the Path of the Vertex, on the aforesaid Plane: and this is the Projection we are to delineate. This being presupposed, it will follow;

1. That when the Sun is in ♋︎, ♌︎, ♍︎, ♎︎, ♏︎, ♐︎, the Northern half of the Earth’s Axis projected on the aforesaid Plane, viewed on that Side next to the Earth, lies to the Right-Hand from the Axis of the Ecliptick: But if the Longitude of the Sun be in any of the six opposite Signs, it lies to the Left-Hand from the Axis of the Ecliptick.
2. When the Sun’s apparent Place happens to be either in ♈︎, ♉︎, ♊︎, ♋︎, ♌︎, ♍︎, the North Pole lies in the illuminate, or visible part of the Disk; but otherways in the obscure.
3. When the Sun’s Place in the Ecliptick is 90 Degrees distant from either Pole; that is, when the Sun is in the Equator, the Paths of the Vertices, or all Circles of the Earth parallel to the Equator, will be projected in right Lines upon the said Plane: but if the Sun’s Place be lesser than 90 Degrees, the said Paths will be projected in Ellipses upon the said Plane, whose conjugate Diameters will be so much the lesser, as the Place of the Sun is lesser.
4. The transverse Diameter of the Ellipses representing any Path, is equal to double the right Sine of the Distance of the said Vertex from the Pole; that is, equal to twice the Co-Sine of the Latitude of the Place or Vertex: but the Conjugate, to the Difference of the right Sines of the Sum, and Difference of the Distances of the Path and Sun from the Pole; that is, equal to the Sine of the Complement of the Sun’s Declination added to the Co-Latitude of the Place, less the right Sine of the Difference of the Complement of the Sun’s Declination, and the Co-Latitude of the Place.
5. The transverse Diameter lies at right Angles to the Earth’s Axis, and the conjugate co-incides therewith.

## Section I.

### To represent in Plano, the Path of a Vertex in the Earth’s Disk, whose distance from the North Pole is 38 Deg. 32 Min. the Sun’s place being in 10 Deg. 40 Mi. 30 Sec. of Gemini, semblable to that which will be projected on a Plane, touching the Earth’s Orbit in that Point, by strait Lines produced from the Sun to the Earth.

Having drawn the Semi-Circle HER, let it represent the Northern half of the Earth’s illuminate Disk (because the Sun is in Gemini) projected upon the said Plane, the Sun it’s Center, the Point therein opposite to the Sun, H☉︎R an Arc of the Ecliptick passing through it. Upon ☉︎ raise GE, perpendicular to the Ecliptick HR, and the Point E wherein it intersects the Limb of the Disk, will be the Pole of the Ecliptick, and ☉︎E it’s Axis.

Again; Make ☉︎E equal to the Radius of a Line of Chords (by Use III. of the Line of Chords), from which taking the Chord of 23 Deg. 30 Min. (the constant Distance of the two Poles) set it off both ways from E to B and C, draw the Line BC, in which the Northern Pole of the World shall be found.

Make BA equal to AC, the half of this Line, the Radius of a Line of Sines, and therein set off the Sine of the Sun’s Distance from the solstitial Colure 19 Deg. 19 Min. 30 Sec. from A to P, on the Left-Hand of the Axis of the Ecliptick (because the Sun is in Gemini), and draw the Line OP, which will be the Axis of the Earth, and P the Place of the North-Pole in the illuminate Hemisphere of the Disk.

Or the Angle E☉︎I, which the Axis of the Earth and Ecliptick make with each other, may be more accurately determined by Calculation. For,

 As Radius—to the Sine of the Sun’s Distance from the Solstitial Colure 90° 00′ 00″ 10,000000 19° 19′ 30″ 9,519731 So is the Tangent of the Sun’s greatest Declination to the Tangent of the Inclination of the Axis 23° 30′ 00″ 9,637956 8° 10′ 54″ 9,157687

Count the said 8 Deg. 10 Min. 54 Sec. in the Limb of the Disk from E to I, on the Left-hand, and draw the Line ☉︎I, this shall be the Axis; and the Point P wherein it intersects the Line BC, the Place of the Pole in the illuminate Disk.

The next thing required will be the Sun’s Distance from the Pole, or the Complement of his Declination, which will be found 67 Deg. 57 Min. 48 Sec. this added to the Distance of the Vertex from the Pole 38 Deg. 32 Min. makes 106 Deg. 29 Min. 48 Sec. and the same 38 Deg. 32 Min. taken from 67 Deg. 51 Min. 48 Sec. gives 29 Deg. 25 Min. 48 Sec. the Meridional Distance of the Sun from the Vertex.

Make ☉︎E the Radius of the Disk, to be the Radius of a Line of Sines, from which take the Sine of 73 Deg. 30 Min. 12 Sec. (the Complement of 106 Deg. 29 Min. 48 Sec. to a Semi-Circle) and set it off in the Axis from ☉︎ to 12; it there gives the Meridional Intersection of the Nocturnal Arc of the Path with the Axis.

Take the Sine of 29 Deg. 25 Min. 48 Sec. from the same Line of Sines, and set it off the same way from to M, and it there gives the Intersection of the diurnal Arc of the Path with the Meridian. Whence M12 will be the conjugate Diameter of the Path, it being the Difference of the Sines of 70 Deg. 30 Min. 12 Sec. and 29 Deg. 25 Min. 48 Sec. that is, the Difference of the Sines of the Sum, and Difference of the Distances of the Path and Sun from the Pole, which will be the conjugate Diameter of any Path.

Bisect 12M in C, and through it draw 6C6 at right Angles to the Axis of the Globe; and then taking the Sine of 38 Deg. 32 Min. the Distance of the Pole from the Vertex, set it off from C both ways to 6 and 6; then the Line 66 will be the Tranverse-Diameter of the Path, and C6 the Semi-Diameter.

Making C6 equal to the Radius of a Line of Sines, if from the same you take the right Sines of 15, 30, 45, 60, 75 Degrees, and set them off severally both ways from C in the Tranverse-Diameter, and from the Points so found erect Perpendiculars, a11, a1, a10, a2, &c. equal to the Co-Sines of the said Arcs, taken from a Line of Sines, whose Radius shall be C12, equal to CM, you will have twenty-four Points given, through which the Ellipsis representing the Path shall pass, which shall also shew the Place of the Vertex at every Hour of the Day. In the same manner may the Parts of an Hour be pricked down in the Path, in laying off the Sine of the Degrees and Minutes corresponding thereto from C towards 6, and then raising Perpendiculars from the Points so found in the Semi-Transverse, and setting off from the said Semi-Transverse each way upon the Perpendiculars, the Sines of the Complements of the Degrees and Minutes corresponding to the aforesaid Parts of an Hour. As, for Example; To denote half an Hour past 11 and 12, cake the Sine of 7 Deg. 30 Min. and lay it off on both sides from C to b and b; then take the Co-sine of 7 Deg. 30 Min. and having raised the Perpendiculars b$$\frac{1}{2}$$, lay off the laid Sine-Complement from b to $$\frac{1}{2}$$, and you will have the Points in the Periphery of the Ellipsis, for half an Hour past 11, and half an Hour past 12; and in this manner may the Path be divided into Minutes, if the Ellipsis be large enough.

Take this for another Example; Suppose I would represent upon the Plane of the Earth’s Disk, the Path of Gibraltar, whose Latitude is 35 Deg. 32 Min. North, and the Sun’s Place is in 15 Deg. 45 Min. of Leo.

Having, as before, drawn the Semi-Circle HER, for the Northern half of the Earth’s illuminate Disk, and drawn ☉︎E perpendicular to RH, as also drawn the Lane CB, which is always equal to twice the Chord of the Sun’s greatest Declination, 23 Deg. 30 Min. you must next make AB equal to a Radius of a Line of Sines, and then lay off from A to P, on the Right-hand of the Axis of the Ecliptick (because the Sun is in Leo), the Sine of the Sun’s Distance from the solstitial Colure 45 Deg. 45 Min. or, the Angle E☉︎I may be more nicely determined by Calculation, as was before directed, and then ☉︎PI, will be the Axis of the World.

Now the Sun’s Distance from the Pole, or the Complement of his Declination is 73 Deg. 51 Min. which being added to the Complement of the Latitude 54 Deg. 28 Min. the Sum will be 128 Deg. 19 Min. and this taken from 180 Deg. the Remainder will be 51 Deg. 41 Min. also if 54 Deg. 28 Min. be taken from 73 Deg. 51 Min. the Difference will be 19 Deg. 23 Min.

Then if you make the Semi-Diameter of the Disk the Radius of a Line of Sines, and lay off from the Center ☉︎ to 12, the Sine of 55 Deg. 41 Min. the Point 12 in the Axis will be the Meridional Intersection of the Nocturnal Arc of the Path with the Axis; and if again you lay off the Sine of 19 Deg. 23 Min. from ☉︎ to M, you will have the Meridional Interaction of the Diurnal Arc of the Path with the Axis; where M12 will be the conjugate Diameter of the Elliptical Path.

And if you bisect M12 in C, and draw the Line 6C6 at right Angles to the Axis ☉︎I; and then lay off the Sine Complement of the Latitude 54 Deg. 28 Min. from C to 6, on each side the Axis, you will have the Transverse diameter of the Path, which may be drawn and divided, as before directed, for that of Fig. I.

Note, When the elevated Pole is in the obscure Hemisphere of the Earth, the diurnal Arc, or illuminated Part of the Path, is in that Part of the Ellipsis that lies nearest to the said Pole, but otherways in the more remote; and where the Ellipsis cuts the Limb of the Disk, are the Points on it from which the Sun appears to rise and set, &c. And because these Points are necessary to be found, when an Eclipse happens near Sun-rising or Sun-setting, they may be determined in the following manner:

Lay off the Sun’s Declination 22 Deg. 2 Min. upon the Limb of the Disk from R to N, as also the Complement of the Latitude of 38 Deg. 32 Min. from R to P; then draw the Line ☉︎N, and from the Point P let fall upon the Diameter RH, the Perpendicular PQ, cutting the Line ☉︎N in L. This being done, take the Extent GL, between your Compasses, and lay it off upon the Axis ☉︎I from ☉︎ to K; then draw a Line both ways from the Point K, parallel to the transverse Axis C6 of the Path, and the said Line will cut the Limb of the Disk in the Points qp of the Sun’s rising and setting.

Or the Arc Ip may be more accurately determined by Calculation; for in the Triangle ☉︎QL, right-angled at Q, are given the Angle QL☉︎, equal to the Sun’s Distance from the Pole; and the Side Q☉︎ equal to the Sine of the Latitude. To find the Side ☉︎L, which is equal to the Sine Complement of the Arc Ip, the Canon is, As the Sine of the Sun’s Distance from the Pole, is to Radius; So is the Sine of the Latitude to the Sine Complement of the Arc Ip, or Iq.

## Section II.

Having in the foregoing Section shewn how to draw the Path of any Vertex upon the Earth’s Disk, as likewise to divide it, the next things necessary to be given, in order to construct the Phases of a Solar Eclipse in any given Place on the Earth’s Superficies, are;

1. The apparent Time of the nearest Approach of the Moon to the Center of the Disk, or the Time of the Middle of the Eclipse.
2. The nearest Distance of the Moon’s Center from the Center of the Disk in her Passage over it; which is equal to her Latitude at the Time of the Conjunction.
3. The Semi-Diameter of the Disk at the Time of the Conjunction.
4. The Moon’s Semi-Diameter at the same Time.
5. The Sun’s Semi-Diameter.
6. The Semi-Diameter of the Penumbra.
7. The Angle of the Moon’s Way with the Ecliptick, which is equal to the Angle that the Perpendicular to the Moon’s Way forms with the Axis of the Ecliptick; and if the Argument of Latitude be more than 9 Sines, or less than 3, the said Perpendicular lies to the Left-hand; if more, to the Right, from the Axis of the Ecliptick.
8. The hourly Motion of the Moon from the Sun at the Time of the Conjunction.

Note, The Semi-Diameter of the Disk is always equal to the Difference of the Sun and Moon’s horizontal Parallaxes.

All these for the Solar Eclipse of May 11, 1724, will be as follows:

 The apparent Time of the nearest Approach of the Moon to the Center of the Disk, will be 5 Hr. 12 Min. 0 Sec. Afternoon The nearest Distance of the Moon’s Center from the Center of the Disk 0 Hr. 32 Min. 14 Sec. The Semi-Diameter of the Disk 0 Hr. 61 Min. 38 Sec. The Moon’s Semi-Diameter 0 Hr. 16 Min. 32 Sec. The Sun’s Semi-Diameter 0 Hr. 15 Min. 53 Sec. The Semi-Diameter of the Penumbra 0 Hr. 32 Min. 35 Sec. The Angle of the Moon’s Way with the Ecliptick 0 Hr. 5 Deg. 37 Min. The hourly Motion of the Moon from the Sun 0 Hr. 35 Min. 18 Sec.

These being found from Astronomical Tables and Calculations, I shall shew how to draw the Line of the Moon’s Way, or Path of the Penumbra, upon the Plane of the Earth’s Disk, as it falls at the Time of the Conjunction of May 11, 1724, and the manner of dividing the same, for London, Genoa, and Rome.

Having drawn the Semi-Circle HER of the Earth’s Disk, and the Paths of London, Genoa, and Rome, by the Directions of the last Section, the Sun’s Place being 61 Deg. 38 Min. 45 Sec. and the Latitude of London 51 Deg. 30 Min. that of Genoa 44 Deg. 27 Min. and that of Rome 41 Deg. 51 Min. you must next draw the Perpendicular to the Moon’s Way; which is done thus: Take the Semi-Diameter ☉︎H of the Disk between your Compasses, and open your Sector so, that the Distance from 60 to 60 of Chords be equal to that Extent; then taking 5 Deg. 37 Min. parallel-wise from the Lines of Chords (which is the Angle of the Moon’s Way with the Ecliptick, or the Angle that a Perpendicular to her Way makes with the Axis E☉︎ of the same Ecliptick), lay them off upon the Limb of the Disk from E to F, on the Right-hand of the Axis of the Ecliptick, because the Argument of Latitude is more than three Sines, and the Line ☉︎F being drawn, will be the Perpendicular to the Moon’s Way at the Time of the general Conjunction, May 11, 1724.

Again: Take the Semi-Diameter of the Disk between your Compasses, and open the Sector so, that the Distance from 61$$\frac{38}{60}$$, the Semi-Diameter of the Disk, on each Line of Lines be equal to that Extent; then the Sector remaining thus opened, take between your Compasses the parallel Extent of 32$$\frac{14}{60}$$, the nearest Approach of the Moon to the Center of the Disk, and lay it off from ☉︎ to M, upon the Perpendicular to the Moon’s Way; then, if upon the Point M, a Perpendicular, as MG, be drawn both ways, this will be the Line of the Moon’s Way, or Path of the Penumbra.

Now to divide the said Path into it's proper Hours, which let be for London. The middle of the general Eclipse, or the Time when the Moon’s Center will be at M, happens at 12 Minutes past 5 in the Afternoon: say, As 1 Hour or 60 Minutes is to 35 Min. 18 Sec. the hourly Motion of the Moon from the Sun; So is 12 Minutes the Time more than 5 in the Afternoon, to 7 Min. 3 Sec. the Motion from 5 a-Clock to the middle.

Your Sector remaining opened to the last Angle it was set to, take the Extent from 7$$\frac{3}{60}$$ to 7$$\frac{3}{60}$$ on each Line of Lines, and setting one Foot of your Compasses upon M, with the other make a Point on the Moon’s Way to the Right-hand; and this shall be the Place of the Penumbra at 5 a-Clock in the Afternoon at London; which therefore denote with the Number V.

The hourly Motion of the Moon from the Sun is 35 Min. 18 Sec. therefore take the parallel Extent of 35$$\frac{18}{60}$$, on the Line of Lines, between your Compasses, and setting one Foot upon V, with the other make Points on each side V, these shall shew the Place of the Moon’s Center at the Hours of IV and VI; and if from these Points you farther set off the said Extent in the said Line, you may thereby find the Place of the Moon’s Center for every Hour, whilst the Penumbra shall touch the Disk: and if the Space between every Flour be divided into 60 equal Parts, you shall have the Place of the Moon’s Center in the Line of her Way, to every single Minute of Time.

Or, you may take the Semi-Diameter of the Disk between your Compasses, and make a Scale thereof, in dividing it, by means of the Sector, in the following manner: Open the Sector so, that the Distance between 61$$\frac{18}{60}$$, the Semi-Diameter of the Disk, and 61$$\frac{36}{60}$$ on the Line of Lines, be equal to the Semi-Diameter of the Disk. This Distance lay off from A to B: then your Sector remaining thus opened, take between your Compasses successively, the parallel Distances of each Division to 61=$$\frac{38}{60}$$, and lay them off from A towards B, every 5th of which Number, and your Scale will be divided into Minutes. And by the same Method you may divide each Minute into Parts, serving for Seconds, if your Scale be long enough. Now your Scale being divided, you may make use thereof, for drawing and dividing the Path of the Penumbra, without the Sector: For 32$$\frac{14}{60}$$ of these Parts of the Scale, give you the nearest Distance of the Moon’s Center to the Center of the Disk. Also 7$$\frac{3}{60}$$ Parts of the said Scale, will be the Distance of the Center of the Penumbra from the Point M, at five a-Clock; and 35$$\frac{18}{60}$$ of the Parts of the Scale, will be the Distance from Hour to Hour, on the Path of the Penumbra.

Now to fix Numbers upon the said Path of the Penumbra, representing the Hours when the Moon’s Center will be at the said Hours, at Rome and Genoa, we must have the Difference of Longitude between London and the said two Places given; as also, whether they are to the East or West from London; the Difference of Longitude between London and Rome, is 12 Deg. 37 Min. and between London and Genoa, is 9 Deg. 37 Min. they being both to the East from London. Each of these being reduced to Time, the former will be 50 Minutes, and the latter 38 Minutes, wherefore 5 a-Clock for Rome on the Moon’s Way, must be at 10. Min. past 4, for London; and 6 a-Clock at 10 Minutes past five, &c. Understand the same for other Hours and Minutes. I have noted the Hours for Rome under the Line of the Moon’s Way, with Roman Characters. Again, 5 a-Clock on the Moon’s Way for Genoa, must be set at 22 Minutes past 5 for London; and 6 a-Clock, at 22 Minutes past 6, &c. I have noted the Hours for Genoa with small Figures over the Line of the Moon’s Way.

Note, The 10 Minutes, and 22, are each of them the Complement of 50 Minutes, and 38 Minutes to 60 Minutes.

## Section III.

### To determine the apparent Time of the Beginning or End of a Solar Eclipse, the Time when the Sun shall be eclipsed to any possible Number of Digits, the Inclination of the Cusps so the Eclipse, and the Time of the visible Conjunction of the Luminaries in any given Latitude.

The Paths of London, Rome, and Genoa, as also the Path of the Penumbra being drawn and divided, as directed in the two last Sections for the great Eclipse of 1724, which will be a very proper Example for sufficiently explaining this Method, take between your Compasses the Semi-Diameter of the Penumbra 32$$\frac{35}{60}$$, from the Line of Lines on the Sector, it being first opened to the Semi-Diameter of the Disk 61$$\frac{33}{60}$$, or you may take it from your Scale, which being done, carry one Foot of your Compasses along the Line of the Moon’s Way, from the Right-hand to the Left; wherein find such a Point, that if the said Foot be set, the other Foot shall cut the same Hour or Minute, in the Path of the Vertex of any given Place; then the Points in the Paths upon which either of the Feet of your Compasses stand, will shew the Time of the Beginning of the Eclipse at that Place.

For example; If you carry the Semi-Diameter of the Disk along the Line of the Moon’s Way, you will find that one Foot of the Compasses being set at a, on the Moon’s Way, which is 41 Min. past 5 in the Afternoon for London, the other Foot will fall on the Point b on the Path of London, which is likewise 41 Min. past 5 in the Afternoon; wherefore the Beginning of the Eclipse at London will be at 41 Min. past 5 in the Afternoon.

Again: If you carry still on the Foot of your Compasses, they remaining yet opened to the Semi-Diameter of the Disk, and find another Point on the Moon’s Way, whereon if you fix one Point of your Compasses, the other shall cut the Path of the Vertex at the same Hour or Minute, which this stands upon in the Line of the Moon’s Way, the Points whereon your Compasses stand in either Path, shall shew the Minute the Eclipse ends.

For example: One Foot of the Compasses being set to g in the Path of the Vertex, which is 29 Min. past 7 in the Afternoon, the other Foot will fall upon the Line of the Moon’s Way, at the same Hour and Minute, viz. 29 Min. past 7; therefore the Eclipse ends at London 29 Min. past 7: but take notice, that the Line of the Moon’s Way should be continued further out beyond 7 a-Clock, that so the Point of the Compasses may fall upon the proper Minute, to wit, 29.

Moreover: If one side of a Square be applied to the Ecliptick HR, and so moved backwards or forwards, until the other side of the said Square cuts the same Hour or Minute in the Path of the Vertex, and Line of the Moon’s Way; this same Hour or Minute will be the Time of the visible Conjunction of the Luminaries at the given Place.

For example; When the perpendicular side of the Square cuts the Path of the Moon’s Way at e, which is 37 Min. past 6, the said side will likewise cut the Path of the Vertex for London at c, which is 37 Min. past 6; therefore the Time of the visible Conjunction of the Luminaries at London will be 37 Min. after 6.

Draw the Line ab, as also the Line ☉︎b; this shall represent the vertical Circle, and the Angle ☉︎ba will be the Angle that the vertical Circle makes with the Line connecting the Centers of the Sun and Moon, at the Beginning of the Eclipse at London.

Draw the Line gm; to wit, join the Points in the Path of the Vertex, and the Path of the Moon’s Way, which shews the End of the Eclipse at London; and the Line ☉︎g; then the Angle ☉︎gm, will be that which the vertical Circle forms with the Line joining the Centers of the Luminaries.

Take the Semi-Diameter of the Sun, viz. 15$$\frac{53}{60}$$ between your Compasses, either from your Sector, opened as before directed, to the Semi-Diameter of the Disk, or from your Scale, and with that upon the Center c (to wit, the Minute in the Path of London, whereat the Time of the visible Conjunction happens) describe a Circle; this Circle shall represent the Sun.

Again; Take the Moon’s Semi-Diameter 16$$\frac{42}{60}$$ from your Sector (remaining opened as before), or your Scale, and upon the Center e (to wit, the Minute in the Path of the Moon’s Way, whereat the true Conjunction happens at London) describe another Circle. This shall cut off from the former Circle so much as the Sun will be eclipsed, at the Time of the visible Conjunction.

From ☉︎ draw the Line ☉︎cv: This shall represent the vertical Circle, and v the vertical Point in the Sun’s Limb, whereby the Position of the Cusps of the Eclipse, in respect of the Perpendicular passing thro’ the Sun’s Center, are plainly and easily had.

Produce dc ’till it intersects the Moon’s Limb in p, then shall pq be the part of the Sun’s Diameter eclipsed, at the Time of the greatest Obscuration at London: And if the Sun’s Diameter be divided into 12 equal Parts, or Digits, you will find pq to be 11$$\frac{66}{100}$$ of those Parts or Digits.

Whence at London,

 The Beginning of the Eclipse, May 11, 1724, at 05 Hr. 41 Min. Aftern. The visible Conjunction of the Luminaries 06 Hr. 37 Min.Digits then 11$$\frac{66}{100}$$ The End 07 Hr. 29 Min. After the same manner, the Beginning of the Eclipse at Genoa will be 06 Hr. 27 Min. Visible Conjunction, or the middle of the Eclipse 07 Hr. 20 Min.

The Sun will there set eclipsed, and the Eclipse will be Total.

 And the Beginning of the Eclipse at Rome is 06 Hr. 42 Min.

The visible Conjunction, or Middle, will there be when the Sun is set, and consequently also the End.

I have, as you see in the Figure, also drawn a fourth Path (or Edinburgh, whose Latitude is 55 Deg. 56 Min. and Longitude about 3 Deg. to the West from London. Wherefore for each Hour in the Moon’s Way for London, you must account 12 Min. more for the same Hour at Edinburgh; that is, for example, 5 a-Clock on the Line of the Moon’s Way for Edinburgh) must stand at 12 Min. past 5 at London. Understand the same for other Hours, &c.

And by proceeding according to the Directions before given, you will find,

At Edinburgh,

 The Beginning of the Eclipse, at 05 Hr. 22 Min. Aftern. The Middle 06 Hr. 20 Min.Digits then 11 The End 07 Hr. 14 Min.

Note, The Path of the Moon’s Way ought to be continued out further to the Left-hand, in order to determine the Time of the End of the Eclipse at Edinburgh.

If you have a mind to know at what Time any possible Number of Digits or Minutes shall be eclipsed at any Place in the Sun's antecedent or consequent Limb; divide the Sun’s Diameter into Digits or Minutes, and cut off the Parts required to be eclipsed from the Semi-Diameter of the Penumbra; then take the remaining part of it between your Compasses, and carrying it along the Line of the Moon’s Way, find the first Point in it, in which placing one Foot, the other will cut the same Hour in the Path of the Place that the fixed Foot stands upon; then the Hour and Minute in either Path upon which the Feet of your Compasses stand, will be the Time of that Obscuration.

As, for example; Suppose it was required to find at what Time 6 Digits or $$\frac{1}{2}$$ of the Sun’s Diameter shall be eclipsed in his antecedent Limb at London: Cut off $$\frac{1}{2}$$ of the Sun’s Semi-Diameter from the Semi-Diameter of the Penumbra, and carrying the Remainder, as directed, you will find, that if one Point of your Compasses be set at 6 Hours 9 Minutes in the Afternoon, on the Path of the Moon’s Way, the other Point will also fall upon the same Hour and Minute in the Path of London; and therefore the Time when the Sun’s antecedent Limb at London will be half eclipsed, will be at 9 Minutes past 6; and when it’s consequent Limb will be half eclipsed, will be at 5 Minutes past 7.

Now to determine the Position of the Cusps of the Eclipse, for example, at London: Draw a Circle ADBE, representing the Sun’s Body, and the right Line ACB, representing his vertical Diameter. This being done, lay off the Angle ☉︎ba upon the Sun’s Limb from A to D, draw the Diameter ECD, and the Point D will be the first Point of the Sun’s Limb obscured by the Moon at the Beginning of the Eclipse.

Again; To determine the Position and Appearance of the Eclipse at the Time of the middle, or greatest Obscuration, take the Sun’s Semi-Diameter between your Compasses, and upon the Point C, describe a Circle; then draw the vertical Diameter ACB, and make the Angle ACD equal to the Angle vcp, and draw the Diameter DCF. This being done take the Moon’s Semi-Diameter between your Compasses, and having laid off from the Center C to E, the Distance ce in. the first Figure; upon the Point E, as a Center, describe an Arc cutting the Sun’s Limb, and the Position and Appearance of the Eclipse at the Time of the greatest Obscuration, or the middle, at London, will be as you see in the Figure.

Lastly, To determine the Position of the End of the Eclipse, draw a Circle (as in the 4th Figure), and cross it with the vertical Diameter ACB; then make the Angle ACE equal to the Angle ☉︎gm, and draw the Diameter ED; then will the Point E on the Limb of the Sun, be that which is last obscured, or whereat the Eclipse ends.

If you have a mind to find the Continuation of total Darkness at any Place where the Sun will be totally eclipsed, cut off the Semi-Diameter of the Sun, from the Semi-Diameter of the Penumbra, and taking the Remainder between your Compasses, carry it along the Line of the Moon’s Way, and find the first Point in it; on which placing one Foot, the other will cut the same Hour in the Path of the Place, which Hour note down. Again; Carrying on further the same Extent of your Compasses, find two Points on the Paths of the Vertex and Moon’s Way, which shall shew the same Hour and Minute on them both. This Time also note down; then substract the Time before found from this Time, and the Difference will be the Time of Continuance of total Darkness.