An essay on the use of celestial and terrestrial globes; exemplified in a greater variety of problems, than are to be found in any other work; exhibiting the general principles of dialing & navigation. / By the late George Adams, mathematical instrument maker to His Majesty, and optician to the Prince of Wales.

PART IV. OF THE USE OF THE CELESTIAL GLOBE.

THE celestial globe is an artificial represen∣tation of the heavens, having the fixed stars drawn upon it, in their natural order and situation; whilst it's rotation on it's axis repre∣sents the apparent diurnal motion of the sun, moon, and stars.

It is not known how early the ancients had any thing of this kind: we are not certain what the sphere of Atlas or Musaeus was; perhaps Palamedes, who lived about the time of the Trojan war, had something of this kind; for of him it is said,

To mark the signs that cloudless skies bestow, To tell the seasons, when to sail and plow, He first devised; each planet's order found, It's distance, period, in the blue profound.

From Pliny it would seem that Hipparchus had a celestial globe with the stars delineated upon it.

It is not to be supposed that the celestial globe is so just a representation of the heavens as the terrestrial globe is of the earth; because here the stars are drawn upon a convex surface, whereas they naturally appear in a concave one. But suppose the globe were made of glass, then to an eye placed in the center, the stars which are drawn upon it would appear in a concave surface, just as they do in the hea∣vens.

Or if the reader was to suppose that holes were made in each star, and an eye placed in the center of the globe, it would view, through those holes, the same stars in the heavens that they represent.

As the terrestrial globe, by turning on it's axis, represents the real diurnal motion of the earth; so the celestial globe, by turning on it's axis, represents the apparent diurnal motion of the heavens.

For the sake of perspicuity, and to avoid continual references, it will be necessary to re∣peat here some articles which have been already mentioned.

The ecliptic is that graduated circle which crosses the equator in an angle of about 23½ de∣grees,

and the angle is called the obliquity of the ecliptic.

This circle is divided into twelve equal parts, consisting of 30 degrees each; the begin∣nings of them are marked with characters, re∣presenting the twelve signs.

Aries ♈, Taurus ♉, Gemini ♊, Cancer ♋, Leo ♌, Virgo ♍, Libra ♎, Scorpio ♏, Sagit∣tarius ♐, Captricornus ♑, Aquarius ♒, Pisces ♓.

Upon my father's globes, just under the ecliptic, the months, and days of each month, are graduated, for the readier fixing the artifi∣cial sun upon it's place in the ecliptic.

The two points where the ecliptic crosses the equinoctial, (the circle that answers to the equator on the terrestrial globe) are called the equinoctial points; they are at the beginnings of Aries and Libra, and are so called, because when the sun is in either of them, the day and night is every where equal.

The first points of Cancer and Capricorn are called solstitial points; because when the sun arrives at either of them, he seems to stand in a manner still for several days, in respect to his distance from the equinoctial; when he is in one solstitial point, he makes to us the longest day; when in the other, the longest night.

The latitude and longitude of stars are de∣termined from the ecliptic.

The longitude of the stars and planets is reckoned upon the ecliptic; the numbers be∣ginning at the first points of Aries ♈, where the ecliptic crosses the equator, and increasing according to the order of the signs.

Thus suppose the sun to be in the 10th de∣gree of Leo, we say, his longitude, or place, is four signs, ten degrees; because he has already passed the four signs, Aries, Taurus, Gemini, Cancer, and is ten degrees in the fifth.

The latitude of the stars and planets is de∣termined by their distance from the ecliptic upon a secondary or great circle passing through it's poles, and crossing it at right an∣gles.

Twenty-four of these circular lines, which cross the ecliptic at right angles, being fifteen degrees from each other, are drawn upon the surface of our celestial globe; which being pro∣duced both ways, those on one side meet in a point on the northern polar circle, and those on the other meet in a point on the southern polar circle.

The points determined by the meeting of these circles are called the poles of the ecliptic, one north, the other south.

From these definitions it follows, that lon∣gitude and latitude, on the celestial globe, bear just the same relation to the ecliptic, as they do on the terrestrial globe to the equator.

Thus as the longitude of places on the earth is measured by degrees upon the equator, count∣ing from the first meridian; so the longitude of the heavenly bodies is measured by degrees upon the ecliptic, counting from the first point of Aries.

And as latitude on the earth is measured by degrees upon the meridian, counting from the equator; so the latitude of the heavenly bodies is measured by degrees upon a circle of longi∣tude, counting either north or south from the ecliptic.

The sun, therefore, has no latitude, being al∣ways in the ecliptic; nor do we usually speak of his longitude, but rather of his place in the ecliptic, expressing it by such a degree and min∣ute of such a sign, as 5 degrees of Taurus, in∣stead of 35 degrees of longitude.

The distance of any heavenly body from the equinoctial, measured upon the meridian, is called it's declination.

Therefore, the sun's declination, north or south, at any time, is the same as the latitude of any place to which he is then vertical, which is never more than 23½ degrees.

Therefore all parallels of declination on the celestial globe are the very same as parallels of latitude on the terrestrial.

Stars may have north latitude and south de∣clination, and vice versa.

That which is called longitude on the ter∣restrial globe, is called right ascension on the celestial; namely, the sun or star's distance from that meridian which passes through the first point of Aries, counted on the equinoctial.

Astronomers also speak of oblique ascension and descension, by which they mean the distance of that point of the equinoctial from the first point of Aries, which in an oblique sphere rises or sets, at the same time that the sun or star rises or sets.

Ascensional difference is the difference be∣twixt right and oblique ascension. The sun's ascensional difference turned into time, is just so much as he rises before or after six o'clock.

The celestial signs and constellations on the surface of the celestial globe, are represented by a variety of human and other figures, to which the stars that are either in or near them, are referred.

The several systems of stars, which are ap∣plied to those images, are called constellations. Twelve of these are represented on the ecliptic circle, and extend both northward and south∣ward from it. So many of those stars as fall within the limits of 8 degrees on both sides of the ecliptic circle, together with such parts of their images as are contained within the afore∣said bounds, constitute a kind of broad hoop▪ belt, or girdle, which is called the zodiac.

The names and the respective characters of the twelve signs of the ecliptic may be learned by inspection on the surface of the broad paper circle, and the constellations from the globe it∣self.

The zodiac is represented by eight circles parallel to the ecliptic, on each side thereof; these circles are one degree distant from each other, so that the whole breadth of the zodiac is 16 degrees.

Amongst these parallels, the latitude of the planets is reckoned; and in their apparent motion they never exceed the limits of the zodiac.

On each side of the zodiac, as was observed, other constellations are distinguished; those on the north side are called northern, and those on the south side of it, southern constellations.

OF THE PRECESSION OF THE EQUINOXES.

All the stars which compose these constella∣tions, are supposed to increase their longitude continually; upon which supposition, the whole starry firmament has a slow motion from west to east; insomuch that the first star in the con∣stellation of Aries, which appeared in the ver∣nal intersection of the equator and ecliptic in the time of Meton the Athenian, upwards of

1900 years ago, is now removed about 30 de∣grees from it.

This change of the stars in longitude, which has now become sufficiently apparent, is owing to a small retrograde motion of the equinoctial points, of about 50 seconds in a year, which is occasioned by the attraction of the sun and moon upon the protuberant matter about the equator. The same cause also occasions a small deviation in the parallelism of the earth's axis, by which it is continually directed towards different points in the heavens, and makes a complete revolution round the ecliptic in about 25,920 years. The former of these motions is called the precession of the equinoxes, the latter the nutation of the earth's axis. In consequence of this shifting of the equinoctial points, an alteration has taken place in the signs of the ecliptic; those stars, which in the infancy of astronomy were in Aries, being now got into Taurus, those of Taurus into Gemini, &c.; so that the stars which rose and set at any parti∣cular seasons of the year, in the times of Hesiod, Eudoxus, and Virgil, will not at present answer the descriptions given of them by those writ∣ers.

PROBLEM I.

To represent the motion of the equinoctial points backwards, or in antecedentia, upon the celestial globe, elevate the north pole so that it's axis may be perpendicular to the plane of the broad paper circle, and the equator will then be in the same plane; let these represent the ecliptic, and then the poles of the globe will also represent those of the ecliptic; the ecliptic line upon the globe will at the same time represent the equator, in∣clined in an angle of 23½ degrees to the broad pa∣per circle, now called the ecliptic, and cutting it in two points, which are called the equinoctial intersections.

Now if you turn the globe slowly round upon it's axis from east to west, while it is in this position, these points of intersection will move round the same way; and the inclination of the circle, which in shewing this motion re∣presents the equinoctial, will not be altered by such a revolution of the intersecting or equi∣noctial points. This motion is called the pre∣cession of the equinoxes, because it carries the equinoctial points backwards amongst the fixed stars.

The poles of the world seem to describe a circle from east to west, round the poles of the ecliptic, arising from the precession of the

equinox. It is a very slow motion, for the equi∣noctial points take up 72 years to move one degree, and therefore they are 25,920 years in describing 360 degrees, or completing a revo∣lution.

This motion of the poles is easily repre∣sented by the above-described position of the globe, in which, if the reader remembers, the broad paper circle represents the ecliptic, and the axis of the globe being perpendicular there∣to, represents the axis of the ecliptic; and the two points, where the circular lines meet, will represent the poles of the world, whence, as the globe is slowly turned from east to west, these points will revolve the same way about the poles of the globe, which are here supposed to represent the poles of the ecliptic. The axis of the world may revolve as above, although it's situation, with respect to the ecliptic, be not altered; for the points here supposed to repre∣sent the poles of the world, will always keep the same distance from the broad paper circle, which represents the ecliptic in this situation of the globe.* 1.1

From the different degrees of brightness in the stars, some appear to be greater than others, or nearer to us; on our celestial globe they are dis∣tinguished into seven different magnitudes.

OF THE USE OF THE CELESTIAL GLOBE, IN THE SOLUTION OF PROBLEMS RELATIVE TO THE SUN.

EVERY thing that relates to the sun is of such importance to man, that in all things he claims a natural preheminence. The sun is at once the most beautiful emblem of the Su∣preme Being, and, under his influence, the fostering parent of worlds; being present to them by his rays, cheering them by his counte∣nance, cherishing them by his heat, adorning them at each returning spring with the gayest and richest attire, illuminating them with his light, and feeding the lamp of life.

To the ancients he was known under a va∣riety of names, each characteristic of his dif∣ferent effects; he was their Hercules, the great deliverer, the restorer of light out of darkness, the dispenser of good, continually labouring for the happiness of a depraved race. He was the Mithra of the Persians, a word derived from love, or mercy, because the whole world is cherished by him, and feels as it were the ef∣fects of his love.

In the sacred scriptures, the original source of all emblematical writings, our Lord is called our sun, and the sun of righteousness; and as there is but one sun in the heavens, so there is but one true God, the maker and redeemer of all things, the light of the understanding, and the life of the soul.

As in scripture our God is spoken of as a shield and buckler, so the sun is characterized by this mark ☉, representing a shield or buck∣ler, the middle point, the umbo, or boss; be∣cause it is love, or life, which alone can protect from fear and death.

His celestial rays, like those of the sun, take their circuit round the earth; there is no corner of it so remote as to be without the reach of their vivifying and penetrating power. As the material light is always ready to run it's heavenly race, and daily issues forth with renewed vigour, like an invincible champion, still fresh to labour; so likewise did our re∣deeming God rejoice to run his glorious race, he excelled in strength, and triumphed, and continues to triumph over all the powers of darkness, and is ever manifesting himself as the deliverer, the protector, the friend, and father, of the human race.* 1.2

PROBLEM II.

To rectify the celestial globe.

To rectify the celestial globe, is to put it in that position in which it may represent exactly the apparent motion of the heavens.

In different places, the position will vary, and that according to the different latitude of the places. Therefore, to rectify for any place, find first, by the terrestrial globe, the latitude of that place.

The latitude of the place being found in degrees, elevate the pole of the celestial globe the same number of degrees and minutes above the plane of the horizon, for this is the name given to the broad paper circle, in the use of the celestial globe.

Thus the latitude of London being 51½ de∣grees, let the globe be moved till the plane of the horizon cuts the meridian in that point.

The next rectification is for the sun's place, which may be performed as directed in prob. xxix; or look for the day of the month close under the ecliptic line, against which is the sun's place, place the artificial sun over that point, then bring the sun's place to the gra∣duated edge of the strong brazen meridian, and set the hour index to the most elevated twelve.

Thus on the 24th of May the sun is in 3½ degrees of Gemini, and is situated near the Bull's eye and the seven stars, which are not then visible, on account of his superior light. If the sun were on that day to suffer a total eclipse, these stars would then be seen shining with their accustomed brightness.

Lastly, set the meridian of the globe north and south, by the compass.

And the globe will be rectified, or put into a similar position, to the concave surface of the heavens, for the given latitude.

PROBLEM III.

To find the right ascension and declination of the sun for any day.

Bring the sun's place in the ecliptic for the given day to the meridian, and the degree of the meridian directly over it is the sun's declination for that day at noon. The point of the equinoctial cut by the meridian, when the sun's place is under it, will be the right ascension.

Thus April 19, the sun's declination is 11° 14′ north, his right ascension 27° 30′. On the 1st of December the sun's declination is 21° 54′ south, right ascension 247° 50′.

PROBLEM IV.

To find the sun's oblique ascension and descension, it's eastern and western amplitude, and time of rising and setting, on any given time, in any given place.

1. Rectify the globe for the latitude, the zenith, and the sun's place. 2. Bring the sun's place to the eastern side of the horizon; then the number of degrees intercepted be∣tween a degree of the equinoctial at the hori∣zon and the beginning of Aries, is the sun's oblique ascension. 3. The number of degrees on the horizon intercepted between the east point and the sun's place, is the eastern of rising amplitude. 4. The hour shewn by the index is the time of sun-rising. 5. Carry the sun to the western side of the horizon, and you in the same manner obtain the oblique descension, western amplitude, and time of setting. Thus at London, May 1,

• The sun's oblique ascension 18° 48′
• Eastern amplitude 24 57 N
• Time of rising 4h 40m
• Oblique descension 257° 7′
• Western amplitude 26 9
• Time of setting 7h 4m

PROBLEM V.

To find the sun's meridian altitude.

Rectify the globe for the latitude, zenith, and sun's place; and when the sun's place is in the meridian, the degrees between that point and the horizon are it's meridian altitude. Thus, on May 17, at London, the meridian altitude of the sun is 57° 55′.

PROBLEM VI.

To find the length of any day in the year, in any lat∣itude, not exceeding 66½ degrees.

Elevate the celestial globe to the latitude, and set the center of the artificial sun to his place upon the ecliptic line on the globe for the given day, and bring it's center to the strong brass meridian, placing the horary in∣dex to that XII which is most elevated; then turn the globe till the artificial sun cuts the eastern edge of the horizon, and the horary index will shew the time of sun-rising; turn it to the western side, and you obtain the hour of sun-setting.

The length of the day and night will be obtained by doubling the time of sun-rising and setting, as before.

PROBLEM VII.

To find the length of the longest and shortest days in any latitude that does not exceed 66½ degrees.

Elevate the globe according to the latitude, and place the center of the artificial sun for the longest day upon the first point of Cancer, but for the shortest day upon the first point of Ca∣pricorn; then proceed as in the last problem.

But if the place hath south latitude, the sun is in the first point of Capricorn on their longest day, and in the first point of Cancer on their shortest day.

PROBLEM VIII.

To find the latitude of a place, in which it's long∣est day may be of any given length between twelve and twenty-four hours.

Set the artificial sun to the first point of Cancer, bring its center to the strong brass meridian, and set the horary index to XII; turn the globe till it points to half the number of the given hours and minutes; then elevate or depress the pole till the artificial sun coin∣cides with the horizon, and that elevation of the pole is the latitude required.

PROBLEM IX.

To find the time of the sun's rising and setting, the length of the day and night, on any place whose latitude lies between the polar circles; and also the length of the shortest day in any of those latitudes, and in what climate they are.

Rectify the globe to the latitude of the given place, and bring the artificial sun to his place in the ecliptic for the given day of the month; and then bring it's center under the strong brass meridian, and set the horary index to that XII which is most elevated.

Then bring the center of the artificial sun to the eastern part of the broad paper circle, which in this case represents the horizon, and the horary index shews the time of the sun-rising; turn the artificial sun to the western side, and the horary index will shew the time of the sun-setting.

Double the time of sun-rising is the length of the night, and the double of that of sun-set∣ting is the length of the day.

Thus, on the 5th day of June, the sun rises at 3 h. 40 min. and sets at 8 h. 20 min.; by doubling each number it will appear, that the length of this day is 16 h. 40 min. and that of the night 7 h. 20 min.

The longest day at all places in north lati∣tude, is when the sun is in the first point of Can∣cer. And,

The longest day to those in south latitude, is when the sun is in the first point of Capri∣corn.

Wherefore, the globe being rectified as above, and the artificial sun placed to the first point of Cancer, and brought to the eastern edge of the broad paper circle, and the horary index being set to that XII which is most ele∣vated, on turning the globe from east to west, until the artificial sun coincides with the wes∣tern edge, the number of hours counted, which are passed over by the horary index, is the length of the longest day; their complement to twenty-four hours gives the length of the short∣est night.

If twelve hours be subtracted from the length of the longest day, and the remaining hours doubled, you obtain the climate mentioned by ancient historians; and if you take half the climate, and add thereto twelve hours, you obtain the length of the longest day in that cli∣mate. This holds good for every climate be∣tween the polar circles.

A climate is a space upon the surface of the earth, contained between two parallels of latitude, so far distant from each other▪ that

the longest day in one, differs half an hour from the longest day in the other parallel.

PROBLEM X.

The latitude of a place being given in one of the po∣lar circles, (suppose the northern) to find what number of days (of 24 hours each) the sun doth constantly shine upon the same, how long he is absent, and also the first and last day of his appearance.

Having rectified the globe according to the latitude, turn it about until some point in the first quadrant of the ecliptic (because the latitude is north) intersects the meridian in the north point of the horizon; and right against that point of the ecliptic, on the horizon, stands the day of the month when the longest day be∣gins.

And if the globe be turned about till some point in the second quadrant of the ecliptic cuts the meridian in the same point of the ho∣rizon, it will shew the sun's place when the longest day ends, whence the day of the month may be found, as before; then the number of natural days contained between the times the longest day begins and ends, is the length of the longest day required.

Again, turn the globe about, until some

point in the third quadrant of the ecliptic cuts the meridian in the south part of the horizon; that point of the ecliptic will give the time when the longest night begins.

Lastly, turn the globe about, until some point in the fourth quadrant of the ecliptic cuts the meridian in the south point of the ho∣rizon; and that point of the ecliptic will be the place of the sun when the longest night ends.

Or, the time when the longest day or night begins being known, their end may be found by counting the number of days from that time to the succeeding solstice; then counting the same number of days from the solstitial day, will give the time when it ends.

OF THE EQUATION OF TIME.

It is not possible, in a treatise of this kind, to enter into a disquisition of the nature of time. It is sufficient to observe, that if we would with exactness estimate the quantity of any portion of infinite duration, or convey an idea of the same to others, we make use of such known measures as have been originally borrowed from the motions of the heavenly bodies. It is true, none of these motions are exactly equal and uniform, but are subject to

some small irregularities, which, though of no consequence in the affairs of civil life, must be taken into the account in astronomical calcula∣tions. There are other irregularities of more im∣portance, one of which is in the inequality of the natural day.

It is a consideration that cannot be re∣flected upon without surprise, that wherever we look for commensurabilities and equalities in nature, we are always disappointed. The earth is spherical, but not perfectly so; the summer is unequal, when compared with the winter; the ecliptic disagrees with the equator, and never cuts it twice in the same equinoctial point. The orbit of the earth has an eccen∣tricity more than double in proportion to the spheroidity of it's globe; no number of the re∣volutions of the moon coincides with any num∣ber of the revolutions of the earth in it's or∣bit; no two of the planets measure one another: put thus it is wherever we turn our thoughts, so different are the views of the Creator from our narrow conception of things; where we look for commensuration, we find variety and infinity.

Thus ancient astronomers looked upon the motion of the sun to be sufficiently regular for the mensuration of time; but, by the accurate observations of later astronomers, it is found

that neither the days, nor even the hours, as measured by the sun's apparent motion, are of an equal length, on two accounts.

1st, A natural or solar day of 24 hours, is that space of time the sun takes up in passing from any particular meridian to the same again; but one revolution of the earth, with respect to a fixed star, is performed in 23 hours, 56 min∣utes, 4 seconds; therefore the unequal progres∣sion of the earth through her elliptical orbit, (as she takes almost eight days more to run through the northern half of the ecliptic, than she does to pass through the southern) is the reason that the length of the day is not exactly equal to the time in which the earth performs it's rotation about it's axis.

2dly, From the obliquity of the ecliptic to the equator, on which last we measure time; and as equal portions of one do not correspond to equal portions of the other, the apparent motion of the sun would not be uniform; or, in other words, those points of the equator which come to the meridian, with the place of the sun on different days, would not be at equal distances from each other.

PROBLEM XI.

To illustrate, by the globe, so much of the equation of time as is in consequence of the sun's ap∣parent motion in the ecliptic.

Bring every tenth degree of the ecliptic to the graduated side of the strong brass meridian, and you will find that each tenth degree on the equator will not come thither with it; but in the following order from ♈ to ♋, every tenth degree of the ecliptic comes sooner to the strong brass meridian than their corresponding tenths on the equator; those in the second quadrant of the ecliptic, from ♋ to ♎, come later, form ♎ to ♑ sooner, and from ♑ to Aries later, whilst those at the beginning of each quadrant come to the meridian at the same time; therefore the sun and clock would be equal at these four times, if the sun was not longer in passing through one half of the ecliptic than the other, and the two inequalities joined together, compose that difference which is cal∣led the equation of time.

These causes are independent of each other, sometimes they agree, and at other times are contrary to one another.

The inequality of the natural day is the cause that clocks or watches are sometimes be∣fore, sometimes behind the sun.

A good and well-regulated clock goes uni∣formly on throughout the year, so as to mark the equal hours of a natural day, of a mean length; a sun-dial marks the hours of every day in such a manner, that every hour is a 24th part of the time between the noon of that day, and the noon of the day immediately follow∣ing. The time measured by a clock is called equal or true time, that measured by the sun∣dial apparent time.

THE USE OF THE CELESTIAL GLOBE, IN PRO∣BLEMS RELATIVE TO THE FIXED STARS.

The use of the celestial globe is in no in∣stance more conspicuous, than in the problems concerning the fixed stars. Among many other advantages, it will, if joined with observations on the stars themselves, render the practice and theory of other problems easy and clear to the pupil, and vastly facilitate his progress in astro∣nomical knowledge.

The heavens are as much studded over with stars in the day, as in the night; only they are then rendered invisible to us by the bright∣ness of the solar rays. But when this glorious luminary descends below the horizon, they be∣gin gradually to appear; when the sun is about twelve degrees below the horizon, stars of the first magnitude become visible; when he is

thirteen degrees, those of the second are seen; when fourteen degrees, those of the third mag∣nitude appear; when fifteen degrees, those of the fourth present themselves to view; when he is descended about eighteen degrees, the stars of the fifth and sixth magnitude, and those that are still smaller, become conspicuous, and the azure arch sparkles with all it's glory.

PROBLEM XII.

To find the right ascension and declination of any given star.

Bring the given star to the meridian, and the degree under which it lies is it's decli∣nation; and the point in which the meridian intersects the equinoctial is it's right ascension. Thus the right ascension of Sirius is 99°, it's declination 16° 25′ south; the right ascension of Arcturus is 211° 32′, it's declination 20° 20′ north.

The declination is used to find the latitude of places; the right ascension is used to find the time at which a star or planet comes to the me∣ridian; to find at any given time how long it will be before any celestial body comes to the meridian; to determine in what order those bodies pass the meridian; and to make a cata∣logue of the fixed stars.

PROBLEM XIII.

To find the latitude and longitude of a given star.

Bring the pole of the ecliptic to the meri∣dian, over which fix the quadrant of altitude, and, holding the globe very steady, move the quadrant to lie over the given star, and the de∣gree on the quadrant cut by the star, is it's la∣titude; the degree of the ecliptic cut at the same time by the quadrant, is the longitude of the star.

Thus the latitude of Arcturus is 30° 30′; it's longitude 20° 20′ of Libra: the latitude of Capella is 22° 22′ north; it's longitude 18 8′ of Gemini.

The latitude and longitude of stars is used to fix precisely their place on the globe, to refer planets and comets to the stars, and, lastly, to determine whether they have any mo∣tion, whether any stars vanish, or new ones ap∣pear.

PROBLEM XIV.

The right ascension and declination of a star being given, to find it's place on the globe.

Turn the globe till the meridian cuts the equinoctial in the degree of right ascension▪

Thus, for example, suppose the right ascension of Aldebaran to be 65° 30′, and it's declination to be 16° north, then turn the globe about till the meridian cuts the equinoctial in 65° 30′, and under the 16° of the meridian, on the nor∣thern part, you will observe the star Aldebaran, or the bull's eye.

PROBLEM XV.

To find at what hour any known star passes the me∣ridian, at any given day.

Find the sun's place for that day in the ecliptic, and bring it to the strong brass meri∣dian, set the horary index to XII o'clock, then turn the globe till the star comes to the meri∣dian, and the index will mark the time. Thus on the 15th of August, Lyra comes to the me∣ridian at 45 min. past VIII in the evening. On the 14th of September the brightest of the Pleiades will be on the meridian at IV in the morning.

This problem is useful for directing when to look for any star on the meridian, in order to find the latitude of a place, to adjust a clock, &c.

PROBLEM XVI.

To find on what day a given star will come to the meridian, at any given hour.

Bring the given star to the meridian, and set the index to the proposed hour; then turn the globe till the index points XII at noon, and observe the degree of the ecliptic then at the meridian; this is the sun's place, the day answering to which may be found on the calen∣dar of the broad paper circle.

By knowing whether the hour be in the morning or afternoon, it will be easy to per∣ceive which way to turn the globe, that the proper XII may be pointed to; the globe must be turned towards the west, if the given hour be in the morning, towards the east if it be after∣noon.

Thus Arcturus will be on the meridian at III in the morning on March the 5th, and Cor Leonis at VIII in the evening on April the 21st.

PROBLEM XVII.

To represent the face of the heavens on the globe for a given hour on any day of the year, and learn to distinguish the visible fixed stars.

Rectify the globe to the given latitude of the place and day of the month, setting it due

north and south by the needle; then turn the globe on it's axis till the index points to the given hour of the night; then all the upper hemisphere of the globe will represent the vi∣sible face of the heavens for that time, by which it will be easily seen what constellations and stars of note are then above our horizon, and what position they have with respect to the points of the compass. In this case, supposing the eye was placed in the center of the globe, and holes were pierced through the centers of the stars on it's surface, the eye would perceive through those holes the various corresponding stars in the firmament; and hence it would be easy to know the various constellations at sight, and to be able to call all the stars by their names.

Observe some star that you know, as one of the pointers in the Great Bear, or Sirius; find the same on the globe, and take notice of the position of the contiguous stars in the same or an adjoining constellation; direct your sight to the heavens, and you will see those stars in the same situation. Thus you may proceed from one constellation to another, till you are acquainted with most of the principal stars.

For example: the situation of the stars at London on the 9th of February, at 2 min. past IX in the evening, is as follows.

Sirius, or the Dog-star, is on the meridian,

it's altitude 22°: Procyon, or the little Dog-star, 16′ towards the east, it's altitude 43½: about 24° above this last, and something more towards the east, are the twins, Castor and Pollux: S. 65° E. and 35° in height, is the bright star Regulus, or Cor Leonis: exactly in the east and 22° high, is the star Deneb Alased in the Lion's tail: 30° from the east towards the north Arcturus is about 3° above the horizon: directly over Arc∣turus, and 31° above the horizon, is Cor Caroli: in the north-east are the stars in the extremity of the Great Bear's tail, Aleath the first star in the tail, and Dubhe the northernmost pointer in the same constellation; the altitude of the first of these is 30½, that of the second 41°, and that of the third 56°.

Reckoning westward, we see the beauti∣ful constellation Orion; the middle star of the three in his belt, is S. 20° W. it's altitude 35°: nine degrees below the belt, and a little more to the west, is Rigel the bright star in his heel: above his belt in a strait line drawn from Rigel between the middle and most northward in his belt, and 9° above it, is the bright star in his shoulder: S. 49° W. and 45½ above the horizon, is Aldebaran the southern eye of the Bull: a little to the west of Aldebaran, are the Hyades: the same altitude, and about S. 70° W, are the Pleiades: in the W. by S. point is Capella in Auriga, it's altitude 73°: in the north-west, and

about 42° high, is the constellation Cassiopeia: and almost in the north, near the horizon, is the constellation Cygnus.* 1.3

PROBLEM XVIII.

To trace the circles of the sphere in the starry fir∣mament.

I shall solve this problem for the time of the autumnal equinox; because that intersec∣tion of the equator and ecliptic will be directly under the depressed part of the meridian about midnight; and then the opposite intersection will be elevated above the horizon; and also because our first meridian upon the terrestrial globe passing through London, and the first point of Aries, when both globes are rectified to the latitude of London, and to the sun's place, and the first point of Aries is brought under the graduated side of each of their meri∣dians, we shall have the corresponding face of the heavens and the earth represented, as they are with respect to each other at that time, and the principal circles of each sphere will corres∣pond with each other.

The horizon is then distinguished, if we begin from the north, and count westward, by the following constellations; the hounds and waist of Bootes, the northern crown, the head

of Hercules, the shoulders of Serpentarius, and Sobieski's shield; it passes a little below the feet of Antinous, and through those of Capri∣corn, through the Sculptor's frame, Eridanus, the star Rigel in Orion's foot, the head of Mo∣noceros, the Crab, the head of the Little Lion, and lower part of the Great Bear.

The meridian is then represented by the equinoctial colure, which passes through the star marked δ in the tail of the Little Bear, under the north pole, the pole star, one of the stars in the back of Cassiopeia's chair marked β, the head of Andromeda, the bright star in the wing of Pegasus marked γ, and the extremity of the tail of the Whale.

That part of the equator which is then above the horizon, is distinguished on the western side by the northern part of Sobieski's shield, the shoulder of Antinous, the head and vessel of Aquarius, the belly of the western fish in Pisces; it passes through the head of the Whale, and a bright star marked δ in the corner of his mouth, and thence through the star marked δ in the belt of Orion, at that time near the eastern side of the horizon.

That half of the ecliptic which is then a∣bove the horizon, if we begin from the western side, presents to our view Capricornus, Aqua∣rius, Pisces, Aries, Taurus, Gemini, and a part of the constellation Cancer.

The solstitia colure, from the western side, passes through Cerberus, and the hand of Her∣cules, thence by the western side of the constel∣lation Lyra, and through the Dragon's head and body, through the pole point under the polar star, to the east of Auriga, through the star marked η in the foot of Castor, and through the hand and elbow of Orion.

The northern polar circle, from that part of the meridian under the elevated pole, ad∣vancing towards the west, passes through the shoulder of the Great Bear, thence a little to the north of the star marked α in the Dragon's tail, the great knot of the dragon, the middle of the body of Cepheus, the northern part of Cassiopeia, and base of her throne, through Cameloparda∣lus, and the head of the Great Bear.

The tropic of Cancer, from the western edge of the horizon, passes under the arm of Hercules, under the Vulture, through the Goose and Fox, which is under the beak and wing of the SWAN, under the star called Sheat, marked β in Pegasus, under the head of Andromeda, and through the star marked Φ in the fish of the con∣stellation Pisces, above the bright star in the head of the Ram marked α, through the Pleiades, between the horns of the Bull, and through a group of stars at the foot of Castor, thence above a star marked δ, between Castor and Pol∣lux, and so through a part of the constellation

Cancer, where it disappears by passing under the eastern part of the horizon.

The tropic of Capricorn, from the western side of the horizon, passes through the belly, and under the tail of Capricorn, thence under Aqua∣rius, through a star in Eridanus marked c, thence under the belly of the Whale, through the base of the Chemical Furnace, whence it goes under the Hare at the feet of Orion, being there depres∣sed under the horizon.

The southern polar circle is invisible to the inhabitants of London, by being under our ho∣rizon.

Arctic and antarctic circles, or circles of perpetual apparition and occultation.

The largest parallel of latitude on the terres∣trial globe, as well as the largest circle of decli∣nation on the celestial, that appears entire above the horizon of any place in north latitude, was called by the ancients the arctic circle, or circle of perpetual apparition.

Between the arctic circle and the north pole in the celestial sphere, are contained all those stars which never set at that place, and seem to us, by the rotative motion of the earth, to be perpetually carried round above our horizon, in circles parallel to the equator.

The largest parallel of latitude on the ter∣restrial,

and the largest parallel of declination on the celestial globe, which is entirely hid be∣low the horizon of any place, was by the an∣cients called the antarctic circle, or circle of perpetual occultation.

This circle includes all the stars which never rise in that place to an inhabitant of the northern hemisphere, but are perpetually below the horizon.

All arctic circles touch their horizons in the north point, and all antarctic circles touch their horizons in the south point; which point, in the terrestrial and celestial spheres, is the in∣tersection of the meridian and horizon.

If the elevation of the pole be 45 degrees, the most elevated part either of the arctic or antarctic circle will be in the zenith of the place.

If the pole's elevation be less than 45 de∣grees, the zenith point of those places will fall without it's arctic or antarctic circle; if greater, it will fall within.

Therefore, the nearer any place is to the equator, the less will it's arctic and antarctic circles be; and on the contrary, the farther any place is from the equator, the greater they are. So that,

At the poles, the equator may be con∣sidered as both an arctic and antarctic circle,

because it's plane is coincident with that of the horizon.

But at the equator (that is, in a right sphere) there is neither arctic nor antarctic circle.

They who live under the northern polar circle, have the tropic of Cancer for their arc∣tic, and that of Capricorn for their antarctic circle.

And they who live on either tropic, have one of the polar circles for their arctic, and the other for their antarctic circle.

Hence, whether these circles fall within or without the tropics, their distance from the ze∣nith of any place is ever equal to the difference between the pole's elevation, and that of the equator, above the horizon of that place.

From what has been said, it is plain there may be as many arctic and antarctic circles, as there are individual points upon any one meri∣dian, between the north and south poles of the earth.

Many authors have mistaken these mutable circles, and have given their names to the im∣mutable polar circles, which last are arctic and antarctic circles, in one particular case only, as has been shewn.

PROBLEM XIX.

To find the circle, or parallel of perpetual appa∣rition, or occulation of the fixed star, in a given latitude.

By rectifying the globe to the latitude of the place, and turning it round on it's axis, it will be immediately evident, that the circle of perpe∣tual apparition is that parallel of declination which is equal to the complement of the given latitude northward; and for the perpetual occul∣tation, it is the same parallel southward; that is to say, in other words, all those stars, whose declinations exceed the co-latitude, will al∣ways be visible, or above the horizon; and all those in the opposite hemisphere, whose declina∣tion exceeds the co-latitude, never rise above the horizon.

For instance; in the latitude of London 51 deg. 30 min. whose co-latitude is 38 deg. 30 min. gives the parallels desired; for all those stars which are within the circle, towards the north pole, never descend below our horizon; and all those stars which are within the same circle, about the south pole, can never be seen in the latitude of London, as they never ascend above it's horizon.

OF PROBLEMS RELATING TO THE AZIMUTH, &c. OF THE SUN AND STARS.

PROBLEM XX.

The latitude of the place and the sun's place being given, to find the sun's amplitude.

That degree from east or west in the horizon, wherein any object rises or sets, is called the am∣plitude.

Rectify the globe, and bring the sun's place to the eastern side of the meridian, and the arch of the horizon intercepted between that point and the eastern point, will be the sun's ampli∣tude at rising.

If the same point be brought to the western side of the horizon, the arch of the horizon in∣tercepted between that point and the western point, will be the sun's amplitude at setting.

Thus on the 24th of May the sun rises at four, with 36 degrees of eastern amplitude, that is, 36 degrees from the east towards the north, and sets at eight, with 36 degrees of western am∣plitude.

The amplitude of the sun at rising and set∣ting increases with the latitude of the place: and in very high northern latitudes, the sun scarce sets before he rises again. Homer had

heard something of this, though it is not true of the Laestrygones, to whom he applies it.

Six days and nights an doubtful course we steer; The next, proud LAMOS' lofty towers appear, And Laestygonia's gates arise distinct in air. The shepherd quitting here at night the plain, Calls, to succeed his cares, the watchful swain. But he that scorns the chains of sleep to wear, And adds the herdsman's to the shepherd's care, So near the pastures, and so short the way, His double toils may claim a double pay, And join the labours of the night and day.

PROBLEM XXI.

To find the sun's altitude at any given time of the day.

Set the center of the artificial sun to his place in the ecliptic upon the globe, and rec∣tify it to the latitude and zenith; bring the center of the artificial sun under the strong brass meridian, and set the hour index to that XII which is most elevated; turn the globe to the given hour, and move the graduated edge of the quadrant to the center of the artificial sun; and that degree on the quadrant, which is cut by the sun's center, is the sun's height at that time.

The artificial sun being brought under the strong brass meridian, and the quadrant laid

upon it's center, will shew it's meridian, or greatest altitude, for that day.

If the sun be in the equator, his greatest or meridian altitude is equal to the elevation of the equator, which is always equal to the co∣latitude of the place.

Thus on the 24th of May, at nine o'clock, the sun has 44 deg. altitude, and at six in the afternoon 20 deg.

OF THE AZIMUTHAL OR VERTICAL CIRCLES.

The vertical point, that is, the uppermost point of the celestial globe, represents a point in the heavens, directly over our heads, which is called our zenith.

From this point circular lines may be con∣ceived crossing the horizon at right angles.

These are called azimuth or vertical circles. That one which crosses the horizon at 10 deg. distance, from the meridian on either side, is called an azimuth circle of 10 deg.; that which crosses at 20, is called an azimuth of 20 deg.

The azimuth of 90 deg. is called the prime vertical: it crosses the horizon at the eastern and western points.

Any azimuth circle may be represented by the graduated edge of the brass quadrant of

altitude, when the center upon which it turns is screwed to that point of the strong brass me∣ridian which answers to the latitude of the place, and the place is brought into the ze∣nith.

If the said graduated edge should lie over the sun's center or place, at any given time, it will represent the sun's azimuth at that time.

If the graduated edge be fixed at any point, so as to represent any particular azimuth, and the sun's place be brought there, the horary index will shew at what time of that day the sun will be in that particular azimuth.

Here it may be observed, that the amplitude and azimuth are much the same.

The amplitude shewing the bearing of any object when it rises or sets, from the east and west points of the horizon.

The azimuth the bearing of any object when it is above the horizon, either from the north or south points thereof. These descriptions and illustrations being understood, we may proceed to

PROBLEM XXII.

To find at what time the sun is due east, the day and the latitude being given.

Rectify the globe; then if the latitude and declination are of one kind, bring the quadrant of altitude to the eastern point of the horizon, and the sun's place to the edge of the qua∣drant, and the index will shew the hour.

If the latitude and declination are of dif∣ferent kinds, bring the quadrant to the western point of the horizon, and the point in the ecliptic opposite to the sun's place to the edge of the quadrant, and then the index will shew the hour.

You will easily comprehend the reason of the foregoing distinction, because when the sun is in the equinoctial, it rises due east; but when it is in that part of the ecliptic which is towards the elevated pole, it rises before it is in the eastern vertical circle, and is therefore at that time above the horizon: whereas when it is in the other part of the ecliptic, it passes the eastern prime vertical before it rises, that is below the horizon; whence it is evident, that the opposite point of the ecliptic must then be in the west, and above the horizon. The sun is due east at London at 7 h. 6 min. on the 18th

of May. The second of August, at Cape Horn, the sun is due east at 5 h. 10 min.

PROBLEM XXIII.

To find the rising, setting, and culminating of a star, it's continuance above the horizon, and it's oblique ascension and descension, and also it's eastern and western amplitude, for any given day and place.

I. Rectify the globe to the latitude and ze∣nith, bring the sun's place for the day to the me∣ridian, and set the hour index to XII. 2. Bring the star to the eastern side of the horizon, and it's eastern amplitude, oblique ascension, and time of rising, will be found as taught of the sun. 3. Carry the star to the western side of the horizon; and in the same manner it's wes∣tern amplitude, oblique descension, and time of setting, will be found. 4. The time of ris∣ing, subtracted from that of setting, leaves the continuance of the star above the horizon. 5. This remainder, subtracted from 24 hours, gives the time of it's continuance below the horizon. 6. The hour to which the index points, when the star comes to the meridian, is the time of it's culminating or being on the meridian.

Let the given day be March 14, the place

London the star Sirius; by working the pro∣blem, you will find

It rises at	2 h. 24 min. afternoon.
Culminates at	6 57
Sets at	11 50
Is above the horizon	9 6

It's oblique ascension and descension are 120° 47′, and 77° 15′; it's amplitude 27°, south∣ward.

PROBLEM XXIV.

The latitude, the altitude of the SUN by day, or of a STAR by night, being given, to find the hour of the day, and the sun's or star's azi∣muth.

Rectify the globes for the latitude, the ze∣nith, the sun's place, turn the globe and the quadrant of altitude, so that the sun's place, or the given star, may cut the given degree of altitude, the index will shew the hour, and the quadrant will be the azimuth in the horizon.

Thus on the 21st of August, at London, when the sun's altitude is 36° in the forenoon, the hour is IX, and the azimuth 58° from the south.

At Boston, December 8th, when Rigel had 15 of altitude, the hour was VIII, the azi∣muth S. E. by E. 7°.

PROBLEM XXV.

The latitude and hour of the day being given, to find the altitude and azimuth of the sun, or of a star.

Rectify the globe for the latitude, the ze∣nith, and the sun's place, then the number of degrees contained betwixt the sun's place and the vertex is the sun's meridional zenith dis∣tance; the complement of which to 90 deg. is the sun's meridian altitude. If you turn the globe about until the index points to any other given hour, then bringing the quadrant of al∣titude to cut the sun's place, you will have the sun's altitude at that hour; and where the quadrant cuts the horizon, is the sun's azi∣muth at the same time. Thus May the 1st, at London, the sun's meridian altitude will be 53½ deg.; and at 10 o'clock in the morning, the sun's altitude will be 46 deg. and his azi∣muth about 44 deg. from the south part of the meridian. On the 2d of December, at Rome, at five in the morning, the altitude of Capella is 41 deg. 58 min. it's azimuth 60 deg. 50 min. from N. to W.

PROBLEM XXVI.

The latitude of the place, and the day of the month being given, to find the depression of the sun below the horizon, and the azimuth, at any hour of the night.

Having rectified the globe for the latitude, the zenith, and the sun's place, take a point in the ecliptic exactly opposite to the sun's place, and find the sun's altitude and azimuth, as by the last problem, and these will be the depres∣sion and the altitude required.

Thus if the time given be the 1st of No∣vember, at 10 o'clock at night, the depression and azimuth will be the same as was found in the last problem.

PROBLEM XXVII.

The latitude, the sun's place, and his azimuth being given, to find his altitude, and the hour.

Rectify the globe for the latitude, the ze∣nith, and the sun's place; then put the qua∣drant of altitude to the sun's azimuth in the horizon, and turn the globe till the sun's place meets the edge of the quadrant; then the said edge will shew the altitude, and the index point to the hour.

Thus, May 21st, at London, when the sun

is due east, his altitude will be about 24 deg. and the hour about VII in the morning; and when his azimuth is 60 degrees south-westerly, the altitude will be about 44½ degrees, and the hour 11¾ in the afternoon.

Thus the latitude and the day being known, and having besides either the altitude, the azi∣muth, or the hour, the other two may be easily found.

PROBLEM XXVIII.

The latitude of the place, and the azimuth of the sun or of a star being given, to find the hour of the day or night.

Rectify the globe for the latitude and sun's place, and bring the quadrant of altitude to the given azimuth in the horizon; turn the globe till the sun or star comes to the quadrant, and the index will shew the time. November 5, at Gibraltar, given the sun's azimuth 50 degrees from the south towards the east, the time you will find to be half past VIII in the morning. Given the azimuth of Vega at London, 57 deg. from the north towards the east, February the 8th, the time you will find twenty minutes past II in the morning.

But as it may possibly happen that we may see a star, and would be glad to know what star it is, or whether it may not be a new star, or a

comet, how that may be discovered, will be seen under the following

PROBLEM XXIX.

The latitude of the place, the sun's place, the hour of the night, and the altitude and azi∣muth of any star being given, to find the star.

Rectifying the globe for the latitude of the place, and the sun's place; fix the quadrant of altitude in the zenith, and turn the globe till the hour index points to the given hour, and set the quadrant of altitude to the given azimuth; then the star that cuts the quadrant in the given altitude, will be the star sought.

Though two stars, that have different right ascensions, will not come to the meridian at the same time, yet it is possible that in a certain latitude they may come to the same vertical circle at the same time; and that consideration gives the following

PROBLEM XXX.

The latitude of the place, the sun's place, and two stars, that have the same azimuth, being given, to find the hour of the night.

Rectify the globe for the latitude, the ze∣nith, and the sun's place; then turn the globe,

and also the quadrant about, till both the stars coincide with it's edge; the hour index will shew the hour of the night, and the place where the quadrant cuts the horizon will be the com∣mon azimuth of both stars.

On the 15th of March, at London, the star Betelgeuse, in the shoulder of Orion, and Regel, in the heel of Orion, were observed to have the same azimuth; on working the problem, you will find the time to be 8 hours 47 minutes.

What hath been observed above, of two stars that have the same azimuth, will hold good likewise of two stars that have the same altitude; from whence we have the following

PROBLEM XXXI.

The latitude of the place, the sun's place, and two stars, that have the same altitude, being given, to find the hour of the night.

Rectify the globe for the latitude of the place, the zenith, and the sun's place; turn the globe, so that the same degree on the quadrant shall cut both stars, then the hour index will shew the hour of the night.

In the former propositions, the latitude of the place is supposed to be given, or known; but as it is frequently necessary to find the lati∣tude of the place, especially at sea, how this may be found, in a rude manner at least, hav∣ing

the time given by a good clock, or watch, will be seen in the following

PROBLEM XXXII.

The sun's place, the hour of the night, and two stars, that have the same azimuth, or altitude, being given, to find the latitude of the place.

Rectify the globe for the sun's place, and turn it till the index points to the given hour of the night; keep the globe from turning, and move it up and down in the notches, till the two given stars have the same azimuth, or altitude; then the brass meridian will shew the height of the pole, and consequently the latitude of the place.

PROBLEM XXXIII.

Two stars being given, one on the meridian, and the other on the east and west part of the hori∣zon, to find the latitude of the place.

Bring the star observed on the meridian to the meridian of the globe; then keeping the globe from turning round it's axis, slide the me∣ridian up or down in the notches, till the other star is brought to the east or west part of the ho∣rizon, and that elevation of the pole will be the the latitude of the place sought.

OBSERVATION.

From what hath been said, it appears, that of these five things, 1. the latitude of the place; 2. the sun's place in the ecliptic; 3. the hour of the night; 4. the common azimuth of two known fixed stars; 5. the equal altitude of two known fixed stars; any three of them being given, the remaining two will easily be found.

There are three sorts of risings and settings of the fixed stars, taken notice of by ancient authors, and commonly called poctial risings and settings, because mostly taken notice of by the poets.

These are the cosmical, achronical, and helia∣cal.* 1.4

They are to be found in most authors that treat on the doctrine of the sphere, and are now chiefly useful in comparing and understanding passages in the ancient writers; such are Hesiod, Virgil, Columella, Ovid, Pliny, &c. How they are to be found by calculation, may be seen in Petavius's Uranologion, and Dr. Gregory's Astronomy.

DEFINITION.

When a star rises or sets at sun-rising, it is said to rise or set COSMICALLY.

From whence we shall have the following

PROBLEM XXXIV.

The latitude of the place being given, to find, by the globe, the time of the year when a given star rises or sets cosmically.

Let the given place be Rome, whose lati∣tude is 42 deg. 8 min. north; and let the given star be the Lucida Pleiadum. Rectify the globe for the latitude of the place; bring the star to the edge of the eastern horizon, and mark the point of the ecliptic rising along with it; that will be found to be Taurus, 18 deg. opposite to which, on the horizon, will be found May the 8th. The Lucida Pleiadum, therefore, rises cosmically May the 8th.

If the globe continues rectified as before, and the Lucida Pleiadum be brought to the edge of the western horizon, the point of the ecliptic, which is the sun's place, then rising on the eastern side of the horizon, will be Scor∣pio, 29 deg. opposite to which, on the horizon, will be found November the 21st. The Lu∣cida Pleiadum, therefore, sets cosmically No∣vember the 21st.

In the same manner, in the latitude of Lon∣don, Sirius will be found to rise cosmically August the 10th, and to set cosmically No∣vember the 10th.

It is of the cosmical setting of the Pleiades,

that Virgil is to be understood in this line,

Ante tibi Eooe Atlantides abscondantur,* 1.5

and not of their setting in the cast, as some have imagined, where stars rise, but never set.

DEFINITION.

When a star rises or sets at sun-setting, it is said to rise or set ACHRONICALLY.

Hence, likewise, we have the following

PROBLEM XXXV.

The latitude of the place being given, to find the time of the year when a given star will rise or set achronically.

Let the given place be Athens, whose lati∣tude is 37 deg. north, and let the given star be Arcturus.

Rectify the globe for the latitude of the place, and bringing Arcturus to the eastern side of the horizon, mark the point of the ecliptic then setting on the western side; that will be found Aries, 12 deg. opposite to which, on the horizon, will be found April the 2d. There∣fore Arcturus rises at Athens achronically April the 2d.

It is of this rising of Arcturus that Hesiod speaks in his Opera & Dies.† 1.6

When from the solstice sixty wint'ry days Their turns have finish'd, mark, with glitt'ring rays, From ocean's sacred flood, Arcturus rise, Then first to gild the dusky evening skies.

If the globe continues rectified to the lati∣tude of the place, as before, and Arcturus be brought to the western side of the horizon, the point of the ecliptic setting along with it will be Sagittary, 7 deg. opposite to which, on the horizon, will be found November the 29th. At Athens, therefore, Arcturus sets achroni∣cally November the 29th.

In the same manner Aldebaran, or the Bull's eye, will be found to rise achronically May the 22d, and to set achronically December the 19th.

DEFINITION.

When a star first becomes visible in a morning, after it hath been so near the sun as to be hid by the splendor of his rays, it is said to rise HELIACALLY.

But for this there is required some certain depression of the sun below the horizon, more or less according to the magnitude of the star. A star of the first magnitude is commonly sup∣posed to require that the sun be depressed 12 deg. perpendicularly below the horizon.

This being premised, we have the follow∣ing

PROBLEM XXXVI.

The latitude of the place being given, is find the time of the year when a given star will rise heliacally.

Let the given place be Rome, whose lati∣tude is 42 deg. north, and let the given star be the bright star in the Bull's horn.

Rectify the globe for the latitude of the place, screw on the brass quadrant of altitude in it's zenith, and turn it to the western side of the horizon. Bring the star to the eastern side of the horizon, and mark what degree of the ecliptic is cut by 12 deg. marked on the qua∣drant of altitude; that will be found to be Ca∣pricorn, 3 deg. the point opposite to which is Cancer, 3 deg. and opposite to this will be found on the horizon, June 25th. The bright star, therefore, in the Bull's horn, in the latitude of Rome, rises heliacally June the 25th.

These kinds of risings and settings are not only mentioned by the poets, but likewise by the ancient physicians and historians.

Thus Hippocrates, in his book De AEre, says, "One ought to observe the heliacal risings and settings of the stars, especially the Dog-star, and Arcturus; likewise the cosmical setting of the Pleiades."

And Polybius, speaking of the loss of the

Roman fleet, in the first Punic war, says, "It was not so much owing to fortune, as to the obstinacy of the consuls, in not hearkening to their pilots, who dissuaded them from putting to sea, at that season of the year, which was between the rising of Orion and the Dog-star; it being always dangerous, and subject to storms."* 1.7

DEFINITION.

When a star is first immersed in the evening, or hid by the sun's rays, it is said to set HELI∣ACALLY.

And this again is said to be, when a star of the first magnitude comes within twelve de∣grees of the sun, reckoned in the perpendi∣cular.

Hence again we have the following

PROBLEM XXXVII.

The latitude of the place being given, to find the time of the year when a given star sets heli∣acally.

Let the given place be Rome, in latitude 42 deg. north, and let the given star be the bright star in the Bull's horn. Rectify the globe for the latitude of the place, and bring the star

to the edge of the western horizon; turn the quadrant of altitude, till 12 deg. cut the ecliptic on the eastern side of the meridian. This will be found to be 7 deg. of Sagittary, the point oppo∣site to which, in the ecliptic, is 7 deg. of Ge∣mini; and opposite to that, on the horizon, is May the 28th, the time of the year when that sets heliacally in the latitude of Rome.

OF THE CORRESPONDENCE OF THE CELESTIAL AND TERRESTRIAL SPHERES.

That the reader may thoroughly understand what is meant by the correspondence between the two spheres, let him imagine the celestial globe to be delineated upon glass, or any other transparent matter, which shall invest or sur∣round the terrestrial globe, but in such a man∣ner, that either may be turned about upon the poles of the globe, while the other remains fixed; and suppose the first point of Aries, on the investing globe, to be placed on the first point of Aries on the terrestrial globe, (which point is in the meridian of London) then every star in the celestial sphere will be directly over those places to which it is a correspondent. Each star will then have the degree of it's right ascension directly upon the corresponding de∣gree of terrestrial longitude; their declination

will also be the same with the latitude of the places to which they answer; or, in other words, when the declination of a star is equal to the la∣titude of a place, such star, within the space 24 hours, will pass vertically over that place and all others that have the same latitude.

If we conceive the celestial investing globe to be fixed, and the terrestrial globe to be gra∣dually turned from west to east, it is clear, that as the meridian of London passes from one de∣gree to another under the investing sphere, every star in the celestial sphere becomes cor∣respondent to another place upon the earth, and so on, until the earth has completed one diur∣nal revolution; or till all the stars, by their ap∣parent daily motion, have passed over every meridian of the terrestrial globe. From this view of the subject, an amazing variety, uniting in wonderful and astonishing harmony, pre∣sents itself to the attentive reader; and future ages will find it difficult to investigate the rea∣sons that should induce the present race of astron∣omers to neglect a subject so highly interest∣ing to science, even in a practical view, but which in theory would lead them into more sublime speculations, than any that ever yet presented themselves to their minds.

A GENERAL DESCRIPTION OF THE PASSAGE OF THE STAR MARKED γ IN THE HEAD OF THE CONSTELLATION DRACO, OVER THE PARALLEL OF LONDON.

The star γ, in the head of the constellation Draco, having 51 deg. 32 min. north declina∣tion, equal to the latitude of London, is the correspondent star thereto. To find the places which it passes over, bring London to the gra∣duated side of the brass meridian, and you will find that the degree of the meridian over Lon∣don, and the representative of the star, passes over from London, the road to Bristol, crosses the Severn, the Bristol channel, the counties of Cork and Kerry in Ireland, the north part of the Atlantic ocean, the streights of Belleisle, New Britain, the north part of the province of Canada, New South Wales, the southern part of Kamschatka, thence over different Tartarian nations, several provinces of Russia, over Po∣land, part of Germany, the southern part of the United Provinces, when, crossing the sea, it arrives again at the meridian of London.

When the said star, or any other star, is on the meridian of London, or any other meri∣dian, all other stars, according to their declina∣tion and right ascension, and difference of right ascension, (which answers to terrestrial latitude,

longitude, and difference of longitude) will at the same time be on such meridians, and ver∣tical to such places as correspond in latitude, longitude, and difference of longitude, with the declination, &c. of the respective stars.* 1.8

From the stars, therefore, thus considered, we attain a copious field of geographical know∣ledge, and may gain a clear idea of the propor∣tionable distances, and real bearings, of remote empires, kingdoms, and provinces, from our own zenith, at the same instant of time; which may be found in the same manner as we found the place to which the sun was vertical at any proposed time.

Many instances of this mode of attaining geographical knowledge, may be found in my father's treatise on the globes.

OF THE USE OF THE CELESTIAL GLOBE, IN PROBLEMS RELATIVE TO THE PLANETS.

The situation of the fixed stars being always the same with respect to one another, they have their proper places assigned to them on the globe.

But to the planets no certain place can be assigned, their situation always varying.

That space in the heavens, within the com∣pass of which the planets appear, is called the zodiac.

The latitude of the planets scarce ever ex∣ceeding 8 degrees, the zodiac is said to reach about 8 degrees on each side the ecliptic.

Upon the celestial globe, on each side of the ecliptic, are drawn eight parallel circles, at the distance of one degree from each other, in∣cluding a space of 16 degrees; these are crossed at right angles, with segments of great circles at every 5th degree of the ecliptic; by these, the place of a planet on the globe, on any given day, may be ascertained with accuracy.

PROBLEM XXXVIII.

To find the place of any planet upon the globe, and by that means to find it's place in the heavens: also, to find at what hour any planet will rise or set, or be on the meridian, on any day in the year.

Rectify the globe to the latitude and sun's place, then place the planet's longitude and lati∣tude in an ephemeris, and set the graduated edge of the moveable meridian to the given longitude in the ecliptic, and counting so ma∣ny degrees amongst the parallels in the zodiac, either above or below the ecliptic, as her lati∣tude is north or south; and set the center of the

artificial sun to that point, and the centre will represent the place of the planet for that time.

Or fix the quadrant of altitude over the pole of the ecliptic, and holding the globe fast, bring the edge of the quadrant to cut the given degree of longitude on the ecliptic; then seek the given latitude on the quadrant, and the place under it is the point sought.

While the globe moves about it's axis, this point moving along with it will represent the planet's motion in the heavens. If the planet be brought to the eastern side of the horizon, the horary index will shew the time of it's rising. If the artificial sun is above the hori∣zon, the planet will not be visible: when the planet is under the strong brazen meridian, the hour index shews the time it will be on that circle in the heavens: when it is at the western edge, the time of it's setting will be obtained.

PROBLEM XXXIX.

To find directly the planets which are above the horizon at sun-set, upon any given day and latitude.

Find the sun's place for the given day, bring it to the meridian, set the hour index to XII, and elevate the pole for the given lati∣tude: then bring the place of the sun to the western semicircle of the horizon, and observe

what signs are in that part of the ecliptic above the horizon, then cast your eye upon the ephemeris for that month, and you will at once see what planets possess any of those elevated signs; for such will be visible, and sit for obser∣vation on the night of that day.

PROBLEM XL.

To find the right ascension, declination, amplitude, azimuth, altitude, hour of the night, &c. of any given planet, for a day of a month and la∣titude given.

Rectify the globe for the given latitude and day of the month; then find the planet's place, as before directed, and then the right ascension, declination, amplitude, azimuth, altitude, hour, &c. are all found, as directed in the problems for the sun; there being no differ∣ence in the process, no repetition can be ne∣cessary.

OF THE USE OF THE CELESTIAL GLOBE, IN PROBLEMS RELATIVE TO THE MOON.

From the sun and planets we now proceed to those problems that concern the moon, the brilliant satellite of our earth, which every month enriches it with it's presence; by the mildness of it's light softening the darkness of

night; by it's influence affecting the tide; and by the variety of it's aspects, offering to our view some very remarkable phenomena.

"Soon as the ev'ning shades prevail, The moon takes up the wond'rous tale; And nightly to the lift'ning earth, Repeats the story of her birth: Whilst all the stars that round her burn, And all the planets in their turn, Confirm the tidings as they roll, And spread the truth from pole to pole."

As the orbit of the moon is constantly vary∣ing in it's position, and the place of the node always changing, as her motion is even vari∣able in every part of her orbit, the solutions of the problems which relate to her, are not altogether so simple as those which concern the sun.

The moon increases her longitude in the ecliptic every day, about 13 degrees, 10 minutes, by which means she crosses the meridian of any place about 50 minutes later than she did the preceeding day.

Thus if on any day at noon her place (lon∣gitude) be in the 12th degree of Taurus, it will be 13 deg. 10 min. more, or 25 deg. 10 min. in Taurus on the succeeding noon.

It is new moon when the sun and moon

have the same longitude, or are in or near the same point of the ecliptic.

When they have opposite longitudes, or are in opposite points of the ecliptic, it is full moon.

To ascertain the moon's place with accura∣cy, we must recur to an ephemeris; but as even in most ephemerides the moon's place is only shewn at the beginning of each day, or XII o'clock at noon, it becomes necessary to supply by a table this deficiency, and assign thereby her place for any intermediate time.

In the nautical ephemeris, published under the authority of the Board of Longitude, we have the moon's place for noon and midnight, with rules for accurately obtaining any intermediate time; but as this ephemeris may not always be at hand, we shall insert, from Mr. Martin's trea∣tise on the globes, a table for finding the hourly motion of the moon. In order, however, to use this table, it will be necessary first to find the quantity of the moon's diurnal motion in the eclip∣tic, for any given day; for the quantity of the moon's diurnal motion varies from about 11 deg. 46 min. the least, to 15 deg. 16. min. when greatest.

The following tables are calculated from the least of 11 deg. 46 min. to the greatest of 15 deg. 16 min. every column increasing 10 minutes; upon the top of the column is the

quantity of the diurnal motion, and on the side of the table are the 24 hours, by which means it will be easy to find what part of the diurnal motion of the moon answers to any given num∣ber of hours.

Thus suppose the diurnal motion to be 12° 32′, look on the top column for the number nearest to it, which you will find to be 12° 36′, in the sixth column; and under it, against 9 hours, you will find 4 deg. 43 min. which is her motion in the ecliptic in the space of 9 hours for that day. The quantity of the diur∣nal motion for any day is found by taking the difference between it and the preceding day.

Thus let the diurnal motion for the 11th of May, 1787, be required.

SIGNS.	DEG.	MIN.
On the 11th of May her place was 11	2	35
On the 10th of May 10	19	47
The diurnal motion sought	12	48

TABLES

FOR FINDING THE HOURLY MOTION OF THE MOON, AND THEREBY HER TRUE PLACE AT ANY TIME OF THE DAY.

TABLE I.

HOURS.	11 46	11 56	12 6	12 16	12 26	12 36	12 46	12 56	13 6	13 16	13 26
d. m.	d. m.	d. m.	d. m.	d. m.	d. m.	d. m.	d. m.	d. m.	d. m	d. m
1	0 29	0 30	0 30	0 30	0 31	0 31	0 32	0 32	0 33	0 33	0 34
2	0 59	1 0	1 0	1 1	1 2	1 33	1 4	1 5	1 5	1 6	1 43
3	1 28	1 20	1 31	1 32	1 33	1 35	1 36	1 37	1 38	1 39	1 41
4	1 58	1 59	2 1	2 3	2 4	2 6	2 8	2 9	2 11	2 13	2 14
5	2 27	2 29	2 31	2 34	2 35	2 37	2 40	2 42	2 44	2 46	2 48
6	2 57	2 59	3 1	3 4	3 6	3 9	3 11	3 14	3 16	3 19	3 21
7	3 26	3 29	3 32	3 35	3 38	3 40	3 43	3 46	3 49	3 52	3 55
8	3 55	3 59	4 2	4 6	4 9	4 12	4 15	4 19	4 22	4 25	4 20
9	4 25	4 28	4 32	4 36	4 40	4 43	4 47	4 51	4 55	4 58	5 2
10	4 54	4 58	5 3	5 7	5 11	5 ••	5 19	5 23	5 27	5 32	5 56
11	5 24	5 28	5 33	5 37	5 42	5 ••	5 51	5 56	6 ••	6 3	6 9
12	5 53	5 53	6 3	6 8	6 13	6 18	6 23	6 28	6 33	6 38	6 43
13	6 22	6 28	6 33	6 39	6 44	6 49	6 55	7 ••	7 6	7 11	7 17
14	6 52	6 58	7 3	7 9	7 15	7 21	7 27	7 33	7 38	7 44	7 50
15	7 21	7 27	7 34	7 40	7 46	7 52	7 59	8 5	8 11	8 17	8 24
16	7 51	7 57	8 4	8 11	8 17	8 24	8 31	8 37	8 44	8 51	8 57
17	8 20	8 27	8 34	8 40	8 48	8 55	9 3	9 10	9 17	9 24	9 31
18	8 49	8 ••7	9 4	9 12	9 19	9 27	9 34	9 4••	9 49	9 57	10 4
19	9 19	9 20	9 35	9 43	9 51	9 58	10 6	10 14	10 22	10 30	10 38
20	9 48	9 56	10 5	10 13	10 21	10 30	10 38	10 47	10 55	11 3	11 12
21	10 17	10 26	10 36	10 44	10 53	11 1	11 10	11 19	11 27	11 36	11 43
22	10 47	10 56	11 6	11 15	11 21	11 33	11 42	11 51	12 0	12 10	12 19
23	11 17	11 26	11 36	11 46	11 55	12 4	12 14	12 24	12 33	12 43	12 52
24	11 46	11 56	12 6	12 16	12 26	12 36	12 46	12 56	13 6	13 16	13 26

TABLE II.

HOURS.	13 30	13 40	13 50	14 6	14 16	14 26	14 36	14 46	14 56	15 6	15 16
d. m.	d. m.	d. m.	d. m.	d. m	d. m.	d. m.	d. m.	d. m.	d. m.	d. m
1	0 31	0 34	0 35	0 36	0 36	0 36	0 36	0 37	0 37	0 38	0 38
2	1 8	1 9	1 16	1 10	1 11	1 12	1 13	1 14	1 15	1 15	1 16
3	1 42	1 42	1 46	1 46	1 47	1 48	1 49	1 51	1 51	1 53	1 54
4	2 16	2 8	•• 19	2 21	2 22	2 24	2 26	2 28	2 20	2 31	2 33
5	2 50	2 52	2 54	2 56	2 58	3 0	3 3	3 5	3 7	3 9	3 11
6	3 24	3 26	3 29	3 31	3 34	3 39	3 39	3 41	3 45	3 46	3 9
7	3 58	4 1	4 4	4 7	4 10	4 10	4 15	4 18	4 21	4 24	4 7
8	4 32	4 35	4 39	4 42	4 45	4 49	4 52	4 55	4 59	5 2	5 5
9	5 6	5 10	5 13	5 17	4 21	5 25	5 28	5 32	5 36	5 40	5 43
10	5 40	5 42	5 48	5 52	5 57	6 1	6 5	6 9	6 13	6 17	6 22
11	6 14	6 19	6 23	6 28	6 32	6 37	6 41	6 46	6 51	6 55	7 0
12	6 48	6 53	6 50	7 3	7 8	7 13	7 28	7 23	7 28	7 33	7 28
13	7 22	7 27	7 33	7 38	7 44	7 49	7 54	8 6	8 5	8 11	8 10
14	7 56	8 0	8 8	8 13	8 19	8 25	8 31	8 37	8 43	8 48	8 54
15	8 30	8 36	8 42	8 49	8 55	9 1	9 7	9 14	9 20	9 26	9 32
16	9 4	9 11	9 17	9 21	9 12	9 37	9 44	9 51	9 57	10 4	10 11
17	9 38	9 45	9 52	9 59	10 20	10 13	10 20	10 28	10 33	10 42	10 49
18	10 12	10 19	10 27	10 34	10 42	10 49	10 57	11 4	11 12	11 19	11 27
19	10 46	10 54	11 5	11 10	11 18	11 26	11 34	11 41	11 49	11 57	12 5
20	11 29	11 38	11 37	11 24	11 8	12 2	12 10	12 18	12 17	12 35	12 42
21	11 58	12 3	12 11	12 20	12 9	12 38	12 40	12 55	13 4	13 13	13 21
22	12 28	12 37	12 46	12 55	13 5	13 14	13 23	13 33	13 41	13 50	13 50
23	13 2	13 12	13 21	13 31	13 43	13 59	13 59	14 9	14 10	14 28	14 38
24	13 36	13 46	13 56	14 6	14 16	14 26	14 36	14 46	14 56	15 6	15 16

The moon's path may be represented on the globe in a very pleasing manner, by tying a silken line over the surface of the globe exactly on the ecliptic; then finding, by an ephemeris, the place of the nodes for the given time, con∣fine the silk at these two points, and at 90 de∣grees distance from them elevate the line about 5¼ deg. from the ecliptic, and depress it as much on the other, and it will then represent the lunar orbit for that day.

PROBLEM XLI.

To find the moon's place in the ecliptic, for any given hour of the day.

First without an ephemeris, only knowing the age of the moon, which may be obrained from every common almanack.

Elevate the north pole of the celestial globe to 90 degrees, and then the equator will be in the plane of, and coincide with the broad paper circle; bring the first point of Aries, marked ♈ on the globe, to the day of the new moon on the said broad paper circle, which answers to the sun's place for that day; and the day of the moon's age will stand against the sign and degree of the moon's mean place; to which place apply a small patch to represent the moon.

But if you are provided with an ephemeris,* 1.9 that will give the moon's latitude and place in the ecliptic; first note her place in the ecliptic upon the globe, and then counting so many de∣grees amongst the parallels in the zodiac, either above or below the ecliptic, as her latitude is north or south upon the given day, and that will be the point which represents the true place of the moon for that time, to which apply the ar∣tificial sun, or a small patch.

Thus on the 11th of May, 1787, she was at noon in 2 deg. 35 min. of Pisces, and her lat∣itude was 4 deg. 18 min.; but as her diurnal motion for that day is 12 48 in nine hours, she will have passed over 4 deg. 47 min. which added to her place at noon, gives 7h. 22 min. for her place on the 11th of May, at nine at night.

PROBLEM XLII.

To find the moon's declination for any given day or hour.

The place in her orbit being found, by prob. xli, bring it to the brazen meridian; then the arch of the meridian contained between it and the equinoctial, will be the declination sought.

PROBLEM XLIII.

To find the moon's greatest and least meridian al∣titudes in any given latitude, that of London for example.

It is evident, this can happen only when the ascending node of the moon is in the vernal e∣quinox; for then her greatest meridian altitude will be 5 deg. greater than that of the sun, and therefore about 67 deg.; also her least meridian altitude will be 5 deg. less than that of the sun, and therefore only 10 deg.: there will there∣fore be 57 deg. difference in the meridian alti∣tude of the moon; whereas that of the sun is but 47 deg.

N. B. When the same ascending node is in the autumnal equinox, then will her meridian altitude differ by only 37 deg.; but this pheno∣menon can separately happen but once in the revolution of a node, or once in the space of nineteen years: and it will be a pleasant enter∣tainment to place the silken line to cross the ecliptic in the equinoctial points alternately; for then the reason will more evidently appear, why you observe the moon sometimes within 23 deg. of our zenith, and at other times not more than 10 deg. above the horizon, when she is full south.

PROBLEM XLIV.

To illustrate, by the globe, the phenomenon of the harvest moon.

About the time of the autumnal equinox, when the moon is at or near the full, she is ob∣served to rise almost at the same time for several nights together; and this phenomenon is called the harvest moon.

This circumstance, with which farmers were better acquainted than astronomers, till within these few years, they gratefully ascribed to the goodness of God, not doubting that he had ordered it on purpose to give them an im∣mediate supply of moon-light after sun-set, for their greater convenience in reaping the fruits of the earth.

In this instance of the harvest moon, as in many others discoverable by astronomy, the wisdom and beneficence of the Deity is con∣spicuous, who really so ordered the course of the moon, as to bestow more or less light on all parts of the earth, as their several circum∣stances or seasons render it more or less service∣able.* 1.10

About the equator, where there is no variety of seasons, moon-light is not necessary for ga∣thering in the produce of the ground; and

there the moon rises about 50 minutes later every day or night than on the former. At considerable distances from the equator, where the weather and seasons are more uncertain, the autumnal full moons rise at sun-set from the first to the third quarter. At the poles, where the sun is for half a year absent, the winter full moons shine constantly without set∣ting, from the first to the third quarter.

But this observation is still further con∣firmed, when we consider that this appearance is only peculiar with respect to the full moon, from which only the farmer can derive any ad∣vantage; for in every other month, as well as the three autumnal ones, the moon, for several days together, will vary the time of it's rising very little; but then in the autumnal months this happens about the time when the moon is at the full; in the vernal months, about the time of new moon; in the winter months, about the time of the first quarter; and in the summer months, about the time of the last quarter.

These phenomena depend upon the diffe∣rent angles made by the horizon, and different parts of the moon's orbit, and that the moon can be full but once or twice in a year, in those parts of her orbit which rise with the least an∣gles.

The moon's motion is so nearly in the

ecliptic, that we may consider her at present as moving in it.

The different parts of the ecliptic, on ac∣count of it's obliquity to the earth's axis, make very different angles with the horizon as they rise or set. Those parts, or signs, which rise with the smallest angles, set with the greatest, and vice versa. In equal times, whenever this angle is least, a greater portion of the ecliptic rises, than when the angle is larger.

This may be seen by elevating the globe to any considerable latitude, and then turning it round it's axis in the horizon.

When the moon, therefore, is in those signs which rise or set with the smallest angles, she will rise or set with the least difference of time; and with the greatest difference in those signs which rise or set with the greatest angles.

Thus in the latitude of London, at the time of the vernal equinox, when the sun is setting in the western part of the horizon, the ecliptic then makes an angle of 6•• deg. with the hori∣zon; but when the sun is in the autumnal equi∣nox, and setting in the same western part of the horizon, the ecliptic makes an angle but of 15 deg. with the horizon; all which is evident by a bare inspection of the globe only.

Again, according to the greater or less in∣clination of the ecliptic to the horizon, so a greater or less degree of motion of the globe

about it's axis will be necessary to cause the same arch of the ecliptic to pass through the horizon; and consequently the time of it's passage will be greater or less, in the same pro∣portion; but this will be best illustrated by an example.

Therefore, suppose the sun in the vernal equinox, rectify the globe for the latitude of London, and the place of the sun; then bring the vernal equinox, or sun's place, to the western edge of the horizon, and the hour index will point precisely to VI; at which time, we will also suppose the moon to be in the au∣tumnal equinox, and consequently at full, and rising exactly at the time of sun-set.

But on the following day, the sun, being advanced scarcely one degree in the ecliptic, will set again very nearly at the same time as before; but the moon will, at a mean rate, in the space of one day, pass over 13 deg. in her orbit; and therefore, when the sun sets in the evening after the equinox, the moon will be below the horizon, and the globe must be turned about till 13 deg. of Libra come up to the edge of the horizon, and then the index will point to 7h. 16 min. the time of the moon's rising, which is an hour and quarter after sun∣set for dark night. The next day following there will be 2½ hours, and so on successively, with an increase of 1¼ hour dark night each

evening respectively, at this season of the year; all owing to the very great angle which the ecliptic makes with the horizon at the time of the moon's rising.

On the other hand, suppose the sun in the autumnal equinox, or beginning of Libra, and the moon opposite to it in the vernal equinox, then the globe (rectified as before) being turned about till the sun's place comes to the western edge of the horizon, the index will point to VI, for the time of the setting, and the rising of the full moon on that equinoctial day. On the following day, the sun will set nearly at the same time; but the moon being advanced (in the 24 hours) 13 deg. in the ecliptic, the globe must be turned about till that arch of the eclip∣tic shall ascend the horizon, which motion of the globe will be very little, as the ecliptic now makes so small an angle with the horizon, as is evident by the index, which now points to VI h.17 min. for the time of the moon's rising on the second day, which is about a quarter of an hour after sun-set. The third day, the moon will rise within half an hour; on the fourth, within three quarters of an hour, and so on; so that it will be near a week before the nights will be an hour without illumination; and in greater latitudes this difference will be still greater, as you will easily find by varying the case, in the practice of this celebrated pro∣blem, on the globe.

This phenomenon varies in different years; the moon's orbit being inclined to the ecliptic about five degrees, and the line of the nodes continually moving retrograde, the inclination of her orbit to the equator will be greater at some seasons than it is at others, which prevents her hastening, in each revolution, with an equal pace.

PROBLEM XLV.

To find what azimuth the moon is upon at any place when it is flood, or high water; and thence the high tide for any day of the moon's age at the same place.

Having observed the hour and minute of high water, about the time of new or full moon, rectify the globe to the latitude and sun's place; find the moon's place and latitude in an ephe∣meris, to which set the artificial moon,* 1.11 and screw the quadrant of altitude in the zenith; turn the globe till the horary index points to the time of flood, and lay the quadrant over the center of the artificial moon, and it will cut the horizon in the point of the compass upon

which the moon was, and the degrees on the horizon contained between the strong brass me∣ridian and the quadrant, will be the moon's azimuth from the south.

To find the time of high water at the same place.

Rectify the globe to the latitude and zenith, find the moon's place by an ephemeris for the given day of her age, or day of the month, and set the artificial moon to that place in the zodi∣ac; put the quadrant of altitude to the azimuth before found, and turn the globe till the artifi∣cial moon is under it's graduated edge, and the horary index will point to the time of the day on which it will be high water.

THE USE OF THE CELESTIAL GLOBE IN THE SOLUTION OF PROBLEMS ASCERTAINING THE PLACES AND VISIBLE MOTIONS OR ORBITS OF COMETS.* 1.12

There is another class or species of planets, which are called comets. These move round the sun in regular and stated periods of times, in the same manner, and from the same cause, as the rest of the planets do; that is, by a cen∣tripetal force, every where decreasing as the

squares of the distances increase, which is the general law of the whole planetary system. But this centripetal force in the comets being compounded with the projectile force, in a very different ratio from that which is found in the planets, causes their orbits to be much more elliptical than those of the planets, which are almost circular.

But whatever may be the form of a comet's orbit in reality, their geocentric motions, or the apparent paths which they describe in the heavens among the fixed stars, will always be circular, and therefore may be shewn upon the surface of a celestial globe, as well as the mo∣tions and places of any of the rest of the planets.

To give an instance of the cometary praxis on the globe, we shall chuse that comet, for the subject of these problems, which made it's ap∣pearance at Boston, in New England, in the months of October and November, 1758, in it's return to the sun; after which, it approached so near the sun, as to set heliacally, or to be lost in it's beams for some time spent in passing the perihelion. Then afterwards emerging from the solar rays, it appeared retrograde in it's course from the sun towards the latter end of March, and so continued the whole month of April, and part of May, in the West Indies, particularly in Jamacia, whose latitude ren∣dered

it visible in those parts, when it was, for the greatest part of the time, invisible to us, by reason of it's southern course through the heavens.

When two observations can be made of a comet, it will be very easy to assign it's course, or mark it out upon the surface of the celestial globe. These, with regard to the above-men∣tioned comet, we have, and they are sufficient for our purpose in regard to the solution of cometary problems.

By an observation made at Jamaica on the 31st of March, 1759, at five oclock in the morning, the comet's altitude was found to be 22 deg. 50 min. and it's azimuth 71 deg. south-east. From hence we shall find its place on the surface of the globe by the following pro∣blem.

PROBLEM XLVI.

To rectify the globe for the latitude of the place of observation in Jamaica, latitude 17 deg. 30 min. and given day of the month, viz. March 31st.

Elevate the north pole to 17 deg. 30 min. above the horizon, then fix the quadrant of al∣titude to the same degree in the meridian, or zenith point. Again, the sun's place for the 31st of March is in 10 deg. 34 min. ♈, which

bring to the meridian, and set the hour index at XII, and the globe is then rectified for the place and time of observation.

PROBLEM XLVII.

To determine the place of a comet on the surface of the celestial globe from it's given altitude, azimuth, hour of the day, and latitude of the place.

The globe being rectified to the given la∣titude, and day of the month, turn it about towards the east, till the hour index points to the given time, viz. V o'clock in the morn∣ing; then bring the quadrant of altitude to in∣tersect the horizon in 71 deg. the given azi∣muth in the south-east quarter; then, under 22 deg. 50 min. the given altitude, you will find the comet's place, where you may put a small patch to represent it.

PROBLEM XLVIII.

To find the latitude, longitude, declination, and right ascension of the comets.

In the circles of latitude contained in the zodiac, you will find the latitude of the comet to be about 30 deg. 30 min. from the ecliptic; the same circle of latitude reduces it's place to the ecliptic in 26 deg. 30 min. of ♒, which is

it's longitude sought. Then bring the come∣tary patch to the brazen meridian, and it's de∣clination will be shewn to be 9 deg. 15 min. south. At the same time, it's right ascension will be 227 deg. 30 min.

PROBLEM XLIX.

To shew the time of the comet's rising, southing, setting, and amplitude, for the day of the ob∣servation at Jamacia.

Bring the place of the comet into the eastern semicircle of the horizon, (the globe being rectified as directed) the index will point to III hours 15 min. which is the time of it's rising in the morning at Jamaica, the amplitude 10 deg. very nearly to the south. The patch being brought to the meridian, the index points to IX o'clock 10 min. for the time of culminating, or being south to them. Lastly, bring the patch to touch the western meridian, and the index will point to III in the afternoon, for the time of the comet's setting, with 10 deg. of southern amplitude, of course.

PROBLEM L.

From the comet's place being given, to find the time of it's rising in the horizon of London, on the 31st day of March, 1759.

For this purpose, you need only rectify the globe for the given latitude of London, and bring the cometary patch to the eastern horizon, and the index points to III hours 45 min. for the time of it's rising at London, with about 14 deg. of south amplitude; then turn the patch to the western horizon, and the index points to II hours 25 minutes, the time of it's set∣ting.

N. B. From hence it appears, the comet rose soon enough that morning to have been ob∣served at London, had the heavens been clear, and the astronomers been before-hand appriz∣ed of such a phenomenon.

PROBLEM LI.

To determine another place of the same comet, from an observation made at London on the 6th day of May, at ten in the evening.

On the 6th day of May, 1759, at ten at night, the place of the comet was observed, and it's distance measured with a micrometer, from

two fixed stars marked μ and ν in the constel∣lation called Hydra, and it's altitude was found to be 16 deg. and it's azimuth 37 deg. south-west; from whence it's place on the surface of the globe, is exactly determined, as in prob. xlvii. and having stuck a patch thereon, you will have the two places of the comet on the surface of the globe, for the two distant days and places of observation, as required.

PROBLEM LII.

From two given places of a comet, to assign it's apparent path among the fixed stars in the heavens.

The two places of the comet being deter∣mined by the observations on the 31st of March, 1758, and the 6th of May following, and denoted by two patches respectively, you must move the globe up and down, in the notches of the horizon, till such time as you bring both the patches to coincide with the horizon; then will the arch of the horizon be∣tween the two patches shew, upon the celestial globe, the apparent place of the comet in the interval between the two observations, and by drawing a line with a black lead pencil along by the frame of the horizon, it's path on the surface of the globe will be delineated, as re∣quired. And here it may be observed, that

it's apparent path lay through the following southern constellations, viz. the tail of Capri∣corn, the tail of Piscis Australis, by the head of Indus, the neck and body of Pavo, through the neck of Apus, below Triangulum Australe, above Musca, by the lowermost of the Crosiers, across the hind legs and through the tail of Centaurus, from thence between the two stars in the back of the Hydra before-men∣tioned; after this, it passed on to Sextans Ura∣niae, and then to the ecliptic near Cor Leonis, soon after which it totally disappeared.

PROBLEM LIII.

To estimate the apparent velocity of a comet, two places thereof being given by observation.

Let one place be ascertained near the be∣ginning of it's appearance, and the other to∣wards the end thereof; then bring these two places to the horizon, and count the number of degrees intersected between them, which being the space apparently described in a given time, will be the velocity required. Thus, in the case of the above-mentioned comet, you will find that it described more than 150 deg. in the space of 36 days, which is more than 4 deg per day.

PROBLEM LIV.

To represent the general phenomena of the comet, for any given latitude.

Bring the visible path of the comet to coin∣cide with the horizon, by which it was drawn, and then observe what degree of the meridian is in the north point of the horizon, which, in the case of the foregoing comet, will be the 23 deg. This will shew the greatest latitude in which the whole path can be visible in any latitude less than this, as that of Jamaica; where, for instance, the most southern part of the path will be ele∣vated more than 5 deg. above the horizon, and the comet visible through the whole time of it's apparition. But rectifying the globe for the latitude of London, the path of the said comet will be for the most part invisible, or below the horizon; and therefore it could not have been seen in our latitude, but at times very near the beginning and end of it's appearance; because, by bringing the comet's path on one part to the south point of the horizon, it will immediately appear in what part the comet ceases to be visible; and then bringing the other part of the path to the point, it will appear in what part it will again become visible.

After this manner may the problems relat∣ing to any other comets be performed; and thus the paths of the several comets, which have hitherto been observed, may be severally deli∣neated on the celestial globe, and their various phenomena in different latitudes be thereby shewn.

[figure]

PLATE XIII.

Notes

• * 1.1

  Rutherforth's System of Nat. Philos. vol. ii. p. 730.
• * 1.2

  Horne on the Psalms,
• * 1.3

  Bransby's Use of the Globes.
• * 1.4

  Costard's History of Astronomy.
• * 1.5

  Georg. l. 1. v. 221.
• † 1.6

  Lib. ii. ver. 285.
• * 1.7

  Lib. i. p. 53.
• * 1.8

  Fairman's Geography.
• * 1.9

  The nautical almanack is the best English ephemeris.
• * 1.10

  Ferguson's Astronomy.
• * 1.11

  Or patch representing the moon.
• * 1.12

  Martin's Description and Use of the Globes.