PROBLEM I.
To find the Longitude of any place on the globe.

The reader will find no difficulty in solving this problem, if he recollects the definition we have given of the word longitude, namely, that it is the distance of any place from the first meridian measured on the equator. There­fore, either set the moveable meridian to the place, or bring the place under the strong brass [Page 34] meridian, and that degree of the equator, which is cut by either of the brazen meridians, is the longitude in degrees and minutes, or the hour and minute of its longitude, expressed in time.

As the given place may lie either east or west of the first meridian, the longitude may be expressed accordingly.

It appears most natural to reckon the lon­gitude always westward from the first meri­dian; but it is customary to reckon one half round the globe eastward, the other half west­ward from the first meridian. To accommodate those who may prefer either of these plans, there are two sets of numbers on our globes: the numbers nearest the equator increase west­ward, from the meridian of London quite round the globe to 360°, over which another set of numbers is engraved, which increase the con­trary way; so that the longitude may be reck­oned upon the equator, either east or west.

Example. Bring Boston, in New England, to the graduated edge of either the strong brass, or of the moveable meridian, and you will find it's longitude in degrees to be 70½, or 4 h. 42 min. in time; Rome 12½ degrees east, or 50 min. in time; Charles-Town, North-America, is 79 deg. 50 min. west.

PROBLEM II.
To find the difference of longitude between any two places.

If the pupil understands what is meant by the difference of longitude, the rule for the solution of this problem will naturally occur to his mind. Now the difference of longitude between any two places is the quantity of an angle (at the pole) made by the meridians of those places measured on the equator. To express this angle upon the globe, bring the moveable meridian to one of the places, and the other place under the strong brass circle, and the required angle is contained between these two meridians, the measure or quantity of which is to be counted on the equator.

Example. I find the longitude of Rome to be 12½ east, that of Constantinople to be 29; the difference is 17½ degrees. Again, I find Jerusalem has 35 deg. 25 min. east longitude from London; and Pekin, in China, 116 deg. 52 min. east longitude; the difference is 81 deg. 27 min.; that is, Pekin is 81 deg. 27 min. east longitude from Jerusalem; or Jerusalem is 81 deg. 27 min. west longitude from Pekin.

If one place is east, and the other west of the first meridian, either find the longitude of both places westward, by that set of numbers [Page 36] which increase westward from the meridian of London of 360 deg. and the difference between the number thus found is the answer to the question:—or, add the east and west longitudes, and the sum is the difference of longitude; thus the longitude of Rome is 12 deg. 30 min. east, of Charles-Town 79 deg. 50 min. west; their sum, 91 deg. 20 min. is the difference re­quired.

It may be proper to observe here, that the difference in time is the same with the difference of longitude, consequently that some of the fol­lowing problems are only particular cases of this problem, or readier modes of computing this difference.

PROBLEM III.
To find all those places where it is noon, at any given hour of the day, at any given place.

General rule. Bring the given place to the brass meridian; and set the index to the up­permost XII; then turn the globe till the in­dex points to the given hour, and it will be noon to all the places under the meridian.

At the diurnal motion of the earth is from west to east, it is plain that all places which are to the east of any meridian, must necessarily pass by the sun before a meridian which is to the west can arrive at it.

[Page 37] N. B. As in my father's globes, the XII, or first meridian, passes through London, you have only to bring the given hour to the east of London, if in the morning, to the brass meri­dian, and all those places which are under it will have noon at the given hour; but bring the given hour westward of London, if it be in the afternoon.

When it is 4 h. 50 min. in the afternoon at Paris, it is noon at New Britain, New England, St. Domingo, Terra Firma, Peru, Chili, and Terra del Fuego.

When it is 7 h. 50 min. in the morning at Ispahan, it is noon at the middle of Siberia Chinese Tartary, China, Borneo.

PROBLEM IV.
When it is noon at any place, to find what hour of the day it is at any other place.

Rule. Bring the place at which it is noon, to the strong brass meridian, and set the hour index to the uppermost XII, and then turn the globe about till the other place comes under the strong brass meridian, and the hour index will shew upon the equator the required hour. If to the eastward of the place where it is noon, the hour found will be in the afternoon; if to the westward, it will be in the forenoon.

[Page 38] Thus when it is noon at London, it is 50 min. past XII, at Rome; 32 min. past VII in the evening at Canton, in China; [...]5 min. past VII in the morning at Quebec, in Canada.

PROBLEM V.
The hour being given at any place, to tell what hour it is in any other part of the world.

Rule. Bring the place where the time is required under the strong brass meridian, set the hour index to the given time, then turn the globe, till the other place is under the brass meridian, and the horary index will point to the hour required.

Thus suppose we are at London at IX o'clock in the morning, what is the time at Canton, in China? Answer, 31 min. past IV in the afternoon. When it is IX in the even­ing at London, it is about 15 min. past IV in the afternoon at Quebec, in Canada.

Thus also when it is III in the afternoon at London, it is 18 min. past X in the forenoon at Boston. When it is VI in the morning at the Cape of Good Hope, it is 7 min. after mid­night at Quebec.

PROBLEM VI.
To find the latitude of any place.

If the pupil comprehends the foregoing definition, he will find no difficulty in the so­lution of this and some of the following pro­blems.

Rule. Bring the place to the graduated side of the strong brass meridian, and the de­gree which is over it is the latitude. Thus London will be found to have 51 deg. 30 min. north latitude; Constantinople 41 deg. north latitude; and the Cape of Good Hope 34 deg. south latitude.

PROBLEM VII.
To find all those places which have the same lati­tude with any given place.

Suppose the given place to be London; turn the globe round, and all those places which pass under the same point of the strong brass meridi­an, are in the same latitude.

PROBLEM VIII.
To find the difference of latitude between two places.

Rule. If the places be in the same hemi­sphere, bring each of them to the meridian, and subtract the latitude of one from the other. If they are in different hemispheres, add the lati­tude of one to that of the other.

Example. The latitude of London is 51 deg. 32 min.; that of Constantinople 41 deg.; their difference is 10 deg. 32 min. The differ­ence between London, 51 deg. 32 min. north, and the Cape of Good Hope, 34 deg. south, is 84 deg. 32 min.

PROBLEM IX.
The latitude and longitude of any place being known, to find that place upon the globe.

Rule. Seek for the given longitude in the equator, and bring the moveable meridian to that point; then count from the equator on the meridian, the degree of latitude either to­wards the north of south pole, and bring the artificial horizon to that degree, and the inter­section of it's edge with the meridian is the situ­ation required.

By this problem any place not represented on the globe may be [...] down thereon, and [Page 43] it may be seen where a ship is when it's lati­tude and longitude are known.

Example. The latitude of Smyrna, in Asia, is 38 deg. 28 min. north; it's longitude 27 deg. 30 min. east of London; therefore, bring 27 deg. 30 min. counted eastward on the equator, to the moveable meridian, and slide the diameter of the artificial horizon to 38 deg. 28 min. north latitude, and it's center will be correctly placed over Smyrna.

It may be proper in this place just to shew the pupil, that the latitude of any place is always equal to the elevation of the pole of the same place above the horizon. The reason of this is, that from the equator to the pole are 90 degrees, from the zenith to the horizon are also 90 degrees; the distance of the zenith to the pole is common to both, and therefore if taken away from both, must leave equal remains; that is, the distance from the equator to the zenith, which is the latitude, is equal to the elevation of the pole.

PROBLEM X.
Any place being given, to find the distance of that place from another, in a great circle of the earth.

I shall divide this problem into three cases.

Case 1. If the places lie under the same me­ridian. Bring them up to the meridian, and mark the number of degrees intercepted be­tween them. Multiply the number of degrees thus found by 60, and they will give the num­ber of geographical miles between the two pla­ces. But if we would have the number of En­glish miles, the degrees before found must be multiplied by 69½.

Case 2. If the places lie under the equator. Find their difference of longitude in degrees, and multiply, as in the preceeding case, by 60 or 69½.

Case 3. If the places lie neither under the same meridian, nor under the equator. Then lay the quadrant of altitude over the two places, and mark the number of degrees intercepted between them. These degrees multiplied as above mentioned, will give the required dis­tance.

PROBLEM XI.
To find the angle of position of places.

The angle of position is that formed between the meridian of one of the places, and a great circle passing through the other place.

Rectify the globe to the latitude and zenith of one of the places, bring that place to the strong brass meridian, set the graduated edge of the quadrant to the other place, and the number of degrees contained between it and the strong brass meridian, is the measure of the angle sought. Thus,

The angle of position between the meridian of Cape Clear, in Ireland, and St. Augustine, in Florida, is about 82 degrees south westerly; but the angle of position between St. Augustine and Cape Clear, is only about 46 degrees north easterly.

Hence it is plain, that the line of position, or azimuth, is not the same from either place to the other, as the romb-lines are.

PROBLEM XII.
To find the bearing of one place from another.

The bearing of one sea-port from another is determined by a kind of spiral, called a romb-line, passing from one to the other, so as [Page 55] to make equal angles with all the meridians it passes by; therefore, if both places are situated on the same parallel of latitude, their bearing is either east or west from each other; if they are upon the same meridian, they bear north and south from one another; if they lie upon a romb-line, their bearing is the same with it; if they do not, observe to which romb-line the two places are nearest parallel, and that will shew the bearing sought.

Example. Thus the bearing of the Lizard point from the island of Bermudas is nearly E. N. E.; and that of Bermudas from the Lizard is W. S. W. both nearly upon the same romb-line, but in contrary directions.

PROBLEM XIII.
To place the globe in the same situation, with res­pect to the sun, as our earth is in at the time of the SUMMER SOLSTICE.

Rectify the globe to the extremity of the divisions for the month of June, or 23½ de­grees north declination; that is, bring these di­divisions on the strong brass meridian to coincide with the plane of the broad paper circle.

Then that part of the earth's surface, which is within the northern polar circle, will be above the broad paper circle, and will be in the light, and the inhabitants thereof will have no night.

But all that space which is contained within the southern polar circle, will continue in the shade; that is, it will there be continual night.

In this position of the globe, the pupil will observe how much the diurnal arches of the par­allels of latitude decrease, as they are more and more distant from the elevated pole.

If any place be brought under the strong brass meridian, and the horary index is set to that XII which is most elevated, and the place be afterwards brought to the western side of the broad paper circle, the hour index will shew the time of sun-rising; and when the [Page 62] place is moved to the eastern edge, the index points to the time of sun-setting.

The length of the day is obtained by the time shewn by the horary index, while the globe moves from the west to the east side of the broad paper circle.

Thus it will be found, that at London the sun rises about 15 minutes before IV in the morning, and sets about 15 minutes VIII at night.

At the following places it will be nearly at the times expressed in the table.

We also see, that at the same time the sun is rising at London, it is rising at the isles of Si­cily and Madagascar.

And, that at the same time when the sun sets at London it is setting at the island of Madeira, and at Cape Horn.

And when the sun is setting at the island of Borneo, in the East Indies, it is rising at Flo­rida, in America. And many other similar [Page 63] circumstances relative to other places, are seen as it were by inspection.

PROBLEM XIV.
To explain the situation of the earth, with re­spect to the sun, at the time of the WINTER SOLSTICE.

Rectify the globe to the extremity of the divisions for the month of December, or to 23½ degrees south declination.

When it will be apparent that the whole space within the southern polar circle is in the sun's light, and enjoys continual day; whilst that of the northern polar circle is in the shade, and has continual night.

If the globe be turned round, as before, the horary index will shew, that at the several places before-mentioned their days will be re­spectively equal to what their nights were at the time of the summer solstice.

It will appear farther, that it is now sun­setting at the same time in those places in which it was sun-rising at the same time at the summer solstice; and, on the contrary, sun­rising at the time it then appeared to set.

PROBLEM XV.
To place the globe in the situation of the earth, at the times of the EQUINOX.

The sun has no declination at the times of the equinox, consequently there must be no ele­vation of the pole.

Bring the day of the month when the sun enters the first point of Aries, or day of the month when the sun enters the first point of Libra, to the plane of the broad paper circle; then the two poles of the globe will be in that plane also, and the globe will be in the posi­tion which is called a right sphere.

For it is a right sphere when the two poles are in the plane of the broad paper circle, because then all those circles which are parallel to the equator will be at right angles to that plane.

If the globe be now turned from west to east, it will plainly appear, that all places upon it's surface are twelve hours above the broad paper circle, and twelve hours below it; that is, the days are twelve hours long all over the earth, and the nights are equal to the days, whence these times are called the times of equinox.

Two of these occur in every year; the first [Page 65] is the autumnal, the second the vernal equi­nox.

At these seasons the sun appears to rise at the same time to all places that are on the same meridian. The sun sets also at the same time in all those places.

Thus if London and Mundford, on the gold coast, be brought to the strong brass meridian, the graduated side of which is in this case the horary index, and they be afterwards carried to the western edge of the broad paper circle, the index will shew that the sun rises at VI at both places; when they are carried to the eastern edge, the index points to VI for the time of sun-setting.

N. B. If London be not the given place, the hour index is to be set to the most elevated XII, while the place is under the graduated edge of the strong brass meridian.

The following circumstances, which usually attend the four cardinal divisions of the year, cannot be better introduced than at this place. At the time of the equinoxes, when the sun passes from one hemisphere into the other, there is almost constantly some disturbance in the weather; the winds are then generally higher: at the vernal equinox they are for the most part easterly, cold, dry, and searching. The solstitial point of the summer is often dis­tinguished by violent rains, and what we call [Page 66] a midsummer stood. The winter being less rainy than the summer, nothing particular hap­pens at the winter solstice, but that the frosts commonly set in more severely, with some quantity of snow upon the ground.

PROBLEM XVI.
To exemplify the sun's altitude, as observed with an artificial horizon.

The altitude of the sun is greater or less, according as the line which goes from us to the sun is nearer to, or farther off from our hori­zon.

Let the moveable circle be applied to any place, as London, then will the horizon of Lon­don be thereby represented.

The sun is supposed, as before, to be in the zenith, that is, directly over the terrestrial globe.

If then from London a line go vertically up­wards, the sun will be seen at London in that line.

At sun-rising, when London is brought to the west edge of the broad paper circle, the sup­posed line will be parallel to the artificial hori­zon, and the sun will then be seen in the hori­zon.

As the globe is gradually turned from the west towards the east, the horizon will recede from that line which goes from London verti­cally upwards; so that the line in which the sun is seen gets further and further from the horizon; that is, the sun's altitude increases gradually.

[Page 68] When the horizon, and the line which goes from London vertically upwards, are arrived at the strong brass meridian, the sun is then at his greatest or meridian altitude for that day, and the line and horizon are at the largest angle they can make with each other.

After this, the motion of the globe being continued, the angle between the artificial hori­zon, and the line which goes from London ver­tically upwards, continually decreases, until Lon­don arrives at the eastern edge of the broad pa­per circle; it's horizon then becomes vertical again, and parallel to the line which goes verti­cally upwards. The sun will again appear in the horizon, and will set.

PROBLEM XVII.
Of the sun's meridian altitude, at the three differ­ent seasons.

Rectify the globe to the time of the winter solstice, by problem xiv, and place the center of the visible horizon on London.

When London is at the graduated edge of the strong brass meridian, the line which goes vertically upwards makes an angle of about 15 degrees; this is the sun's meridian altitude at that season, to the inhabitants of London.

If the globe be rectified to the times of equinox, by problem xv, the horizon will be [Page 69] farther separated from the line which goes vertically upwards, and makes a greater angle therewith, it being about 38½ degrees; this is the sun's meridian altitude, at the time of equi­nox at London.

Again, rectify to the summer solstice by problem xiii, and you will find the artificial ho­rizon recede farther from the line which goes from London vertically upwards, and the an­gle it then makes is about 62 degrees, which shews the sun's meridian altitude at the time of the summer solstice.

Hence flows also the following arithmetical problem.

PROBLEM XVIII.
To find the sun's meridian altitude universally.

Add the sun's declination to the elevation of the equator, if the latitude of the place, and the declination of the sun, are both on the same side.

If on contrary sides, subtract the declina­tion from the elevation of the equator, and you obtain the sun's meridian altitude.

PROBLEM XIX.
Of the sun's azimuths, as compared with the artificial horizon.

The artificial horizon serves also to deter­mine the sun's azimuths.

An azimuth of the sun is denominated from that point of the horizon, to which the sun, or a line going to the sun, is nearest.

Thus if the sun, or a line going to the sun, be nearest the south-east point of the horizon, which point is 45 degrees distant from the me­ridian, the sun's azimuth is an azimuth of 45 degrees, and the sun will appear in the south-east.

Imagine the sun, as we have done before, to be placed directly over the globe.

In which case, a line going the sun from any place on the surface of the globe, will have a vertical direction, and will go from that place vertically upwards.

[Page 71] If then we apply the artificial horizon to any place, the point of this horizon to which a vertical line is nearest, shews the sun's azimuth at that time.

It is observable, that the point of the hori­zon to which such a vertical line is nearest, will be at all times that point which is most elevat­ed.

To exemplify this, let the globe be in the position of a right sphere, and let the artificial horizon be applied to London.

When London is at the western edge of the broad paper circle, which situation represents the time when the sun appears to rise, the east­ern point of the artificial horizon being then most elevated, shews that the sun at his rising is due east.

Turn the globe, till London comes to the eastern edge of the broad paper circle, then the western point of the artificial horizon will be most elevated, shewing that the sun sets due west.

Now place the globe in the position of an oblique sphere; and if London be brought to the eastern or western side of the broad paper circle, the vertical line will depart more or less from the east and west points, in which case the sun is said to have more or less ampli­tude

If the departure be northward, it is called [Page 72] northern amplitude; if southward, it is called southern amplitude.

In whatever position the globe be placed,^* when London comes to the strong brass meridi­an, the most elevated part of the artificial hori­zon will be the south point of it.

Which shews that at noon the sun will al­ways, and in all seasons, appear in the south.

PROBLEM XX.
To find the limits of the climates.

Elevate the north pole to 23° 28′, the sun's declination on the longest day; and turn the globe easterly till the intersection of the meridian with the equator that passes through Libra comes to the horizon, and the hour of VI will then be under the meridian, which in this problem is the hour index, because the sun sets this day at places on the equator as it does every day at VI o'clock. Now turn the globe easterly till the time under the meridian is 15 min. past VI. and you find that 8° 34′ of that graduated meridian is cut by the horizon; this is the beginning of the second climate; and the limits of all the cli­mates may be determined, by bringing succes­sively the time equal to half the length of the longest day under the meridian, and observing the degree of the graduated meridian cut by the horizon.

PROBLEM XXI.
To illustrate the distinction of ascii, amphiscii, heteroscii, and periscii, by the globe.

Rectify the globe to the summer solstice, and move the artificial horizon to the equator, the north point will be the most elevated at noon.

Which shews, that to those inhabitants who live at the equator, the sun will at this season appear to the north at noon, and their [Page 79] shadow will therefore be projected south­wards.

But if you rectify the globe to the winter solstice, the south point being then the upper­most point at noon, the same persons will at noon have the sun on the south side of them, and will project their shadows northwards.

Thus they are amphiscii, projecting their shade both ways; which is the case of all the inhabitants within the tropics.

The artificial horizon remaining as before, rectify the globe to the times of the equinox, and you will find that when this horizon is under the strong brass meridian, a line going vertically upwards will be perpendicular to it, and consequently the sun will be directly over the heads of the inhabitants, and they will be ascii, having no noon shade; their shadow is in the morning projected directly westward, in the evening directly eastward.

The same thing will also happen to all the inhabitants who live between the tropics of Cancer and Capricorn; so that they are not only ascii, but amphiscii also.

Those who live without the tropics are heteroscii; those in north latitude have the noon shade always directed to the north, while those in south latitude have it always projected to the south.

The inhabitants of the polar circles are [Page 80] called periscii; because, as the sun goes round them continually, their shade goes round them likewise.

PROBLEM XXII.
To find the Antoeci, the Perioeci, and the Anti­podes of any place.

Bring the given place to the strong brass me­ridian, then in the opposite hemisphere, and un­der the same degree of latitude with the given place, you will find the antoeci.

The given place remaining under the me­ridian, set the horary index to XII; then turn [Page 82] the globe, till the other XII is under the index, then will you find the perioeci under the same de­gree of latitude with the given place.

Thus the inhabitants of the south part of Chili are antoeci to the people of New England, whose Perioeci are those Tartars who dwell on the north borders of China, which Tartars have the said inhabitants of Chili for their an­tipodes.

This will become evident, by placing the globe in the position of a right sphere and bring­ing those nations to the edge of the broad paper circle.

PROBLEM XXIII.
The day of the month being given, to find all those places on the globe, over whose zenith the sun will pass on that day.

Rectify the terrestrial globe, by bringing the given day of the month on the back side of the strong brass meridian, to coincide with the plane of the broad paper circle; observe the number of degrees of the brass meridian, which corresponds to the given day of the month.

This number of degrees, counted from the equator on the strong brass meridian, towards the elevated pole, is the point over which the sun is vertical; and all those places, which pass [Page 83] under this point, have the sun directly vertical on the given day.

Example. Bring the 11th of May to coin­cide with the plane of the broad paper circle, and the said plane will cut eighteen degrees for the elevation of the pole, which is equal to the sun's declination for that day, which being counted on the strong brass meridian towards the elevated pole, is the point over which the sun will be vertical; and all places that are under this degree, will have the sun on their zenith on the 11th of May.

Hence, when the sun's declination is equal to the latitude of any place in the torrid zone, the sun will be vertical to those inhabitants that day; which furnishes us with another me­thod of solving this problem.

PROBLEM XXXV.
To place the terrestrial globe in the sun's rays, that it may represent the natural position of the earth, either by a meridian line, or with­out it.

If you have a meridian line, set the north and south points of the broad paper circle di­rectly over it, the north pole of the globe being elevated to the latitude of the place, and stand­ing upon a level plane, bring the place you are in under the graduated side of the strong brass meridian, then the poles and parallel circles upon the globe will, without sensible error, correspond with those in the heavens, and each [Page 103] point, kingdom, and state, will be turned to­wards the real one which it represents.

If you have no meridian line, then the day of the month being known, find the sun's decli­nation as before instructed, which will direct you to the parallel of the day, amongst the polar parallels, reckoned from either pole to­wards the polar circle; which you are to re­member.

Set the globe upon your horizontal plane in the sun-shine, and put it nearly north and south by the mariner's compass, it being first elevated to the latitude of the place, and the place it­self brought under the graduated side of the strong brass meridian; then move the frame and globe together, till the shade of extuberan­cy, or term of illumination, just touches the polar parallel for the day, and the globe will be settled as before; and if accurately performed, the variation of the magnetic needle will be shewn by the degree to which it points in the compass box.

And here observe, if the parallel for the day should not happen to fall on any one of those drawn upon the globe, you are to esti­mate a proportionable part between them, and reckon that the parallel of the day. If we had drawn more, the globe would have been confused.

The reason of this operation is, that as the [Page 104] sun illuminates half the globe, the shade of ex­tuberancy will constantly be 90 degrees from the point wherein the sun is vertical.

If the sun be in the equator, the shade and il­lumination must terminate in the poles of the world; and when he is in any other diurnal parallel, the terms of illumination must fall short of, or go beyond either pole, as many de­grees as the parallel which the sun describes that day is distant from the equator; therefore, when the shade of extuberancy touches the polar par­allel for the day, the artificial globe will be in the same position, with respect to the sun, as the earth really is, and will be illuminated in the same manner.

PROBLEM XXXVI.
To find naturally the sun's declination, diurnal parallel, and his place thereon.

The globe being set upon an horizontal plane, and adjusted by a meridian line or other­wise, observe upon which, or between which po­lar parallel the term of illumination falls; it's distance from the pole is the degree of the sun's declination; reckon this distance from the equator among the larger parallels, and you have theparallel which the sun describes that day; upon which if you move a card, cut in the form of a double square, until it's shadow [Page 105] falls under itself, you will obtain the very place upon that parallel over which the sun is verti­cal at any hour of that day, if you set the place you are in under the graduated side of the strong brass meridian.

Note, The moon's declination, diurnal pa­rallel, and place, may be found in the same manner. Likewise, when the sun does not shine bright, his declination, &c. may be found by an application in the manner of problem xxxiv.

PROBLEM XXXVII.
To find the sun's azimuth naturally.

If a great circle, at right angles to the ho­rizon, passes through the zenith and nadir, and also through the sun's center, it's distance from the meridian in the morning or evening of any day, reckoned upon the degrees on the inner edge of the broad paper circle, will give the az­imuth required.

PROBLEM XXXVIII.
To shew that in some places of the earth's surface, the sun will be twice on the same azimuth in the morning, twice on the same azimuth in the afternoon: or in other words,

When the declination of the sun exceeds the latitude of any place, on either side of the [Page 107] equator, the sun will be on the same azimuth twice in the morning, and twice in the after­noon.

Thus, suppose the globe rectified to the latitude of Antigua, which is about 17 deg. of north latitude, and the sun to be the begin­ning of Cancer, or to have the greatest north declination; set the quadrant of altitude to the 21st degree north of the east in the horizon, and turn the globe upon it's axis, the sun's center will be on that azimuth at 6h. 30 min. and also at 10h. 30 min. in the morning. At 8h. 30 min. the sun will be as it were sta­tionary, with respect to it's azimuth, for some time; as it will appear by placing the quadrant of altitude to he 17th degree north of the east in the horizon. If the quadrant be set to the same degrees north of the west, the sun's center will cross it twice as it approaches the horizon in the afternoon.

This appearance will happen more or less to all places situated in the torrid zone, when­ever the sun's declination exceeds their lati­tude; and from hence we may infer, that the shadow of a dial, whose gnomon is erected per­pendicular to an horizontal plane, must neces­sarily go back several degrees on the same day.

But as this can only happen within the tor­rid zone, and as Jerusalem lies about 8 degrees [Page 108] to the north of the tropic of Cancer, the retro­cession of the shadow on the dial of Ahaz, at Jerusalem, was, in the strictest signification of the word, miraculous.

PROBLEM XXXIX.
To observe the hour of the day in the most natural manner, when the terrestrial globe is properly placed in the sun-shine.

There are many ways to perform this ope­ration with respect to the hour, three of which are here inserted, being general to all the inha­bitants of the earth; a fourth is added, peculiar to those of London, which will answer, with­out sensible error, at any place not exceeding the distance of 60 miles from this capital.

PROBLEM XL.
To construct an horizontal dial for any given lati­tude, by means of the terrestrial globe.

Elevate the globe to the latitude of the place, then bring the first meridian under the graduated edge of the strong brazen one, which will then be over the hour XII, or the equator. As our globes have meridians drawn through every fifteen degrees of the equator, these me­ridians will represent the true circles of the sphere, and will intersect the horizon of the globe, in certain points on each side of the me­ridian. The distance of these points from the meridian must be carefully noted down upon a [Page 118] piece of paper, as will be seen in the example. The pupil need not, however, take out into his table the distances further than from XII to VI, which is just 90 degrees; for the distances of XI, X, IX, VIII, VII, VI, in the forenoon, are the same from XII as the distances of I, II, III, IV, V, VI, in the afternoon; and these hour-lines continued through the center will give the opposite hour-lines on the other half of the dial.

No more hour-lines need be drawn than what answer to the sun's continuance above the horizon, on the longest day of the year, in the given latitude.

Example. Suppose the given place to be London, whose latitude is 51 deg. 30 min. north.

Elevate the north pole of the globe to 51½ degrees above the horizon; then will the axis of the globe have the same elevation above the broad paper circle, as the gnomon of the dial is to have above the plane thereof.

Turn the globe, till the first meridian (which on English globes passes through Lon­don) is under the graduated side of the strong brazen meridian; then observe and note the points where the hour-circles intersect the ho­rizon; and as on our globes the inner graduated circle, on the broad paper circle, begins from the two sixes, or east and west, we shall begin [Page 119] from thence, calling the hour - VI 0° 0 we shall find the other hours intersecting the horizon at the following degrees:

which are the respective distances of the above hours from VI upon the plane of the horizon.

To transfer these, and the rest of the hours, upon an horizontal plane, draw the parallel right lines a c and b d, fig. 5, plate XIII, upon that plane, as far from each other as is equal to the intended thickness of the gnomon of the dial, and the space included between them will be the meridian, or twelve o'clock line upon the dial; cross this meridian at right angles by the line g h, which will by the six o' clock line; then setting one foot of your compasses in the inter­section a, describe the quadrant g e with any convenient radius, or opening of the compasses; after this, set one foot of the compasses in the in­tersection b, as a center, and with the same rad­ius describe the quadrant f h; then divide each quadrant into 90 equal parts, or degrees, as in the figure.

Because the hour-lines are less distant from each other about noon, than in any other part of the dial, it is best to have the centers of the quadrants at some distance from the center of [Page 120] the dial-plane, in order to enlarge the hour-dis­tances near XII; thus the center of the plane is at A, but the center of the quadrants is at a and b.

Lay a rule over 78′ 9′, and the center b, and draw there the hour-line of I. Through b, and 65 41, gives the hour-line of II. Through b, and 51 57, that of III. Through the same center, and 36 24, we obtain the hour-line of IV. And through it, and 18 54, that of V. And because the sun rises about four in the mor­ning, continue the hour-lines of IV and V in the afternoon, through the center b to the oppo­site side of the dial.

Now lay a rule successively to the center a of the quadrant e g, and the like elevations or degrees of that quadrant, 78 9, 65 41, 51 57, 36 24, 18 54, which will give the forenoon hours of XI, X, IX, VIII, and VII; and be­cause the sun does not set before VIII in the even­ing on the longest days, continue the hour-lines of VII and VIII in the afternoon, and all the hour­lines will be finished on this dial.

Lastly, through 51½ degrees on either quad­rant, and from it's center, draw the right line a g for the axis of the gnomon a g i, and from g let fall the perpendicular g i upon the meridi­an line a i, and there will be a triangle made, whose sides are a g, g i, and i a; if a place simi­lar to this triangle be made as thick as the dis­tance [Page 121] between the lines a c and b d, and be set upright between them, touching at a and b, the line a g will, when it is truly set, be parallel to the axis of the world, and will cast a shadow on the hour of the day.

The trouble of dividing the two quadrants may be saved, by using a line of chords, which is always placed upon every scale belonging to a case of instruments.

PROBLEM XLI.
To delineate a direct south dial for any given lati­tude, by the globe.

Let us suppose a south dial for the latitude of London.

Elevate the pole to the co-latitude of your place, and proceed in all respects as above taught for the horizontal dial, from VI in the morning to VI in the afternoon, only the hours must be reversed, as in fig. 3, plate XIII; and the hy­pothenuse a g of the gnomon a g f, must make an angle with the dial plane to the co-latitude of the place.

As the sun can shine no longer than from VI in the morning to VI in the evening, there is no occasion for having more than twelve hours upon this dial.

In solving this problem, we have considered our vertical south dial for the latitude of Lon­don, [Page 122] as an horizontal one for the complement of that latitude, or 38 deg. 30 min▪; all direct vertical dials may be thus reduced to horizontal ones, in the same manner. The reason of this will be evident, if the globe be elevated to the latitude of London; for by fixing the quadrant of altitude to the zenith, and bringing it to in­tersect the horizon in the east point, it will point out the plane of the proposed dial.

This plane is at right angles to the meri­dian, and perpendicular to the horizon; and it is clear, from the bare inspection of the globe thus elevated, that it's axis forms an angle with this plane, which is just the complement of that which it forms with the horizon, and is there­fore just equal to the co-latitude of the place; and that therefore it is most simple to rectify the globe to that co-latitude.

The north vertical dial is the same with the south, only the stile must point upwards, and that many of the hours from it's direction can be of on use.

PROBLEM XLII.
To make an erect dial, declining from the south towards the east or west.

Elevate the pole to the latitude of the place, and screw the quadrant of altitude to the zenith.

[Page 123] Then if your dial declines towards the east, (which we shall suppose in the present instance) count in the horizon the degrees of declination from the east point towards the north, and bring the lower end of the quadrant to coincide with that degree of declination at which the reckon­ing ends.

Then bring the first meridian under the graduated edge of the strong brass meridian, which strong meridian will be the horary in­dex.

Now turn the globe westward, and observe the degrees cut in the quadrant of altitude by the first meridian, while the hours XI, X, IX, &c. in the forenoon, pass successively under the brazen one; and the degrees thus cut on the quadrant by the first meridian, are the respec­tive distances of the forenoon hours, from XII, on the plane of the quadrant.

For the afternoon hours, turn the quadrant of altitude round the zenith, until it comes to the degree in the horizon, opposite to that where it was placed before, namely, as far from the west towards the south, and turn the globe eastward; and as the hours I, II, III, &c. pass under the strong brazen meridian, the first me­ridian will cut on the quadrant of altitude the number of degrees from the zenith, that each of the hours is from XII on the dial.

When the first meridian goes off the quad­rant [Page 124] at the horizon, in the forenoon, the hour index will shew the time when the sun comes upon this dial; and when it goes off the quad­rant in the afternoon, it points to the time when the sun leaves the dial.

Having thus found all the hour distances from XII, lay them down your dial plane, either by dividing a semicircle into two quad­rants, or by the line of chords.

In all declining dials, the line on which the gnomon stands makes an angle with the twelve o'clock line, and falls among the forenoon hour lines, if the dial declines towards the east; and among the afternoon hour lines, when the dial declines towards the west; that is, to the left hand from the twelve o'clock line in the for­mer case, and to the right hand from it in the latter.

To find the distance of this line from that of twelve.

This may be considered, I. If the dial de­clines from the south towards the east, then count the degrees of that declination in the ho­rizon, from the east point towards the north, and bring the lower end of the quadrant to that degree of declination where the reckon­ing ends; then turn the globe, until the first meridian cuts the horizon in the like number [Page 125] of degrees, counted from the south point to­wards the east, and the quadrant first meri­dian will cross one another at right angles, and the number of degrees of the quadrant, which are intercepted between the first meridian and the zenith, is equal to the distance of this line from the twelve o'clock line.

The numbers of the first meridian, which are intercepted between the quadrant and the north pole, is equal to the elevation of the stile above the plane of the dial.

The second case is, when the dial declines westward from the south.

Count the declination from the east point of the horizon, towards the south, and bring the quadrant of altitude to the degree in the hori­zon, at which the reckoning ends, both for finding the forenoon hours, and the distance of the substile, or gnomon line, from the meridian; and for the afternoon hours, bring the quad­rant to the opposite degrees in the horizon, namely, as far from the west towards the north, and then proceed in all respects as before.

It is presumed, that the foregoing instances will be sufficient to illustrate the general prin­ciples of dialling, and to give the pupil a gene­ral idea of that pleasing science; for accurate and expeditious methods of constructing dials, we must refer him to treatises written expressly on that subject.

PROBLEM XLIII.
Given the difference of latitude, and difference of longitude, to find the course and distance sailed.^*

Example. Admit a ship sails from a port [Page 133] A, in latitude 38 deg. to another port B, in latitude 5 deg. and finds her difference of lon­gitude 43 deg.

Let the port A be brought to the meridian, and elevate the globe to the given latitude of that port 38 deg. and fixing the quadrant of altitude precisely over it on the meridian, move the quadrant to lie over the second port B, (found by the given difference of latitude and longitude) then will it cut in the horizon 50 deg. 45 min. for the angle of the ship's course to be steered from the port A. Also, count the de­grees in the quadrant between the two ports, which you will find 51 deg.; this number mul­tiplied by 60, (the nautical miles in a degree) will give 3060 for the distance run.

PROBLEM XLIV.
Given the difference of latitude and course, to find the difference of longitude and distance sailed.

Example. Admit a ship sails from a port A, in 25 deg. north latitude, on another port B, in 30 deg. south latitude, upon a course of 43 deg.

Bring the port A to the meridian, and rec­tify the globe to the latitude thereof 25 deg. where fix the quadrant of altitude, and place it so as to make an angle with the meridian of [Page 134] 43 deg. in the horizon, and observe where the edge of the quadrant intersects the parallel of 30 deg. south latitude, for that is the place of the port B. Then count the number of de­grees on the edge of the quadrant intersected between the two pieces and there will be found 73 deg. which, multiplied by 6o, gives 4380 miles for the distance sailed. As the two ports are now known, let each be brought to the meridian, and observe the difference of longitude in the equator respectively, which will be found 50 degrees.

N. B. Had this problem been solved by loxodromics, or sailing on a rhumb, the differ­ence of longitude would then have been 52 deg. 30 min. between the two ports.

PROBLEM XLV.
Given the difference of latitude and distance run, to find the deference of longitude, and angle of the course.

Example. Admit a ship sails from a port A, in latitude 50 deg. to another port B, in latitude 17 deg. 30 min. and her distance run be 2220 miles. Rectify the globe to the lati­tude of the place A, then the distance run, re­duced to degrees, will make 37 deg. which are to be reckoned from the end of the quadrant lying over the port A, under the meridian; [Page 135] then is the quadrant be moved, till the 37 deg. coincides with the parallel of 17 deg. 30 min. north latitude; then will the angle of the course appear in the arch of the horizon, inter­cepted between the quadrant and the meri­dian, which will, be 32 deg. 40 min.; and by making a mark on the globe for the port B, and bringing the same to the meridian, you will observe what number of degrees pass under the meridian, which will be 20, the difference of longitude required.

PROBLEM XLVI.
Given the difference of longitude and course, to find the difference of latitude and distance sail­ed.

Example. Suppose a ship sails from A, in the latitude 51 deg. on a course making an angle with the meridian of 40 deg. till the dif­ference of longitude be found just 20 deg.; then rectifying the globe to the latitude of the port A, place the quadrant of altitude so as to make an angle of 40 deg. with the meridian; then observe at what point it intersects the meridian passing through the given longitude of the port B, and there make a mark to repre­sent the said port; then the number of degrees intercepted between that and the port A will be 28, which will give 1680 miles for the dis­tance [Page 136] run. And the said mark for the port B, being brought to the meridian, will have it's latitude there shewn to be 27 deg. 40 min.

PROBLEM XLVII.
Given the course and distance sailed, to find the difference of longitude, and difference of lati­tude.

Example. Suppose a ship sails 1800 miles from a port A, 51 deg. 15 min. south-west, on an angle of 45 deg. to another port B.

Having rectified the globe to the port A, fix the quadrant of altitude over it in the zenith, and place it to the south-west point in the horizon; then upon the edge of the qua­drant under 30 deg.(equal to 1800 miles form the port A) is the port B; which bring to the meridian, and you will there see the latitude; and at the same time, it's longitude on the e­quator, in there point cut by the meridian.

In all these cases, the ship is supposed to be kept upon the arch of a great circle, which is not difficult to be done, very nearly, by means of the globe, by frequently observing the lati­tude, measuring the distance sailed, and (when you can) finding the difference of longitude; for one of these being given, the place and course of the ship is known at the same time; and therefore the preceding course may be al­tered, [Page 137] and rectified without any trouble, through the whole voyage, as often as such observations can be obtained, or it is found necessary. Now if any of these data are but of the quantity of four or five degrees, it will suffice for correcting the ship's course by the globe, and carrying her directly to the intended port, according to the following problem.

PROBLEM XLVIII.
To steer a ship upon the arch of a great circle by the given difference of latitude, or differ­ence of longitude, or distance sailed in a given time.

Admit a ship sails from a port A, to a very distant port Z, whole latitude and longitude are given, as well as it's geographical bearing from A; then,

First, having rectified the globe to the port A, lay the quadrant of altitude over the port Z, and draw thereby the arch of the great circle through A and Z; this will design the intended path or tract of the ship.

Secondly, having kept the ship upon the first given course for some time, suppose by an observation you find the latitude of the present place of the ship, this added to, or subducted from the latitude of the port A, will give the present latitude in the meridian; to which [Page 138] bring the path of the ship, and the part there­in, which lies under the new latitude, is the true place [...] the ship in the great arch. To the latitude of B rectify the globe, and lay the quadrant over Z, and it will shew in the hori­zon the new course to be steered.

Thirdly, suppose the ship to be steered upon this course, till her distance run be found 300 miles, or 5 deg.; then, the globe being recti­fied to the place B in the zenith, laying the quadrant from thence over the great arch, make a mark at the 5th degree from B, and that will be the present place of the ship, which call C; which being brought to the meridian, it's latitude and longitude will be known. Then rectify the globe to the place C, and laying the quadrant from thence to Z, the new course to be steered will appear in the ho­rizon.

Fourthly, having steered some time upon this new course, suppose, by some means or other, you come to know the difference of longitude of the present place of the ship, and of any of the preceding places, C, B, A; as B, for in­stance; then bring B to the meridian, and turn the globe about, till so many degrees of the equator pass under the meridian as are equal to the discovered difference of longitude; then the point of the great arch cut by the me­ridian is the present place D of the ship, to [Page 139] which the new course is to be found as be­fore.

And thus, by repeating these observations at proper intervals, you will find future places, E, F, G, &c. in the great arch; and by recti­fying the course at each, your ship will be conducted on the greatest circle, or the nearest way from the port A to Z, by the use of the globe only.

PROBLEM I.
To find the latitude and longitude of any given place on the globe.

Bring the place to the east side of the brass meridian, then the degree marked the me­ridian over it shews it's latitude, and the degree of the equator under the meridian shews it's lon­gitude.

Hence, if the longitude and latitude of any place be given, the place is easily found, by bringing the given longitude to the meridian; for then the place will lie under the given de­gree of latitude upon the meridian.

PROBLEM II.
To find the difference of longitude between any two given places.

Bring each of the given places successively to the brazen meridian, and see where the me­ridian cuts the equator time; the number of degrees contained between those two points, if it be less than 180 deg. otherwise the remain­der to 360 deg. will be the difference of longi­tude required.

PROBLEM III.
To rectify the globe for the latitude, zenith, and sun's place.

Find the latitude of the place by prob. 1, and if the place be in the northern hemisphere, elevate the north pole above the horizon, ac­cording to the latitude of the place. If the place be in the southern hemisphere, elevate the south pole above the south point of the ho­rizon, as many degrees as are equal to the lat­itude.

Having elevated the globe according to it's latitude, count the degrees thereof upon the meridian from the equator towards the elevat­ed pole, and that point will be the zenith, or the vertex of the place; to this point of the me­ridian fasten the quadrant of altitude, so that the graduated edge thereof may be joined to the said point.

Having brought sun's place in the eclip­tic to the meridian, set the hour index to twelve at noon, and the globe will be rectified to the sun's place.

PROBLEM IV.
The hour of the day at any place being given, to find all those on the globe, where it is noon, midnight, or any given hour at that time.

On the globes when mounted in the com­mon manner, it is the now customary to place the hour-circle under the north pole; it is divided into twice twelve hours, and has two rows of figures, one running from east to west, the other from west to east; this circle is moveable, and the meridian answers the purpose of an index.

Bring the given place to the brazen meri­dian, and the given hour of the day on the hour-circle, this done, turn the globe about, till the meridian points at the hour desired; then, with all those under the meridian, it is noon, midnight, or any given hour at that time.

PROBLEM V.
The hour of the day at any place being given, to find the correspondent hour (or what o'clock it is at that time) in any other place.

Bring the given place to the brazen meri­dian, and set the hour-circle to the given time; then turn the globe about, until the place where the hour is required comes to the [Page 144] meridian, and the meridian will point out the hour of the day at that place.

Thus, when it is noon at London, it is

PROBLEM VI.
The day of the month being given, to find all those places on the globe where the sun will be ver­tical, or in the zenith, that day.

Having found the sun's place in the eclip­tic for the given day, bring the same to the brazen meridian, observe what degree of the meridian is over it, then turn the globe round it's axis, and all places that pass under that degree of the meridian, will have the sun ver­tical, or in the zenith, that day; i. e. directly over the head of each place at it's respective noon.

PROBLEM VII.
A place being given in the torrid zone, to find those two days of the year on which the sun will be vertical to that place.

Bring the given place to the brazen meri­dian, and mark the degree of latitude that is [Page 145] exactly over it on the meridian; then turn the globe about it's axis, and observe the two points of the ecliptic which pass exactly under that de­gree of latitude, and look on the horizon for the two days of the year in which the sun is in those points or degrees of the ecliptic, and they are the days required; for on them, and none else, the sun's declination is equal to the latitude of the given place.

PROBLEM VIII.
To find the antoeci, perioeci, and antipodes of any given place.

Bring the given place to the brazen meri­dian, and having found it's latitude, keep the globe in that position, and count the same number of degrees of latitude on the meridian, from the equator towards the contrary pole, and where the reckoning ends, that will give the place of the antoeci upon the globe. Those who live at the equator have no an­toeci.

The globe remaining in the same position, bring the upper XII on the horary circle to the meridian, then turn the globe about till the me­ridian points to the lower XII; the place which then lies under the meridian, having the same latitude with the given place, is the perioeci re­quired. Those who live at the poles, if any, have no perioeci.

[Page 146] As the globe now stands (with the index at the lower XII), the antipodes of the given place are under the same point of the brazen meri­dian where it's antoeci stood before.

PROBLEM IX.
To find at what hour the sun rises and sets any day in the year, and also upon what point of the compass.

Rectify the globe for the latitude of the place you are in; bring the sun's place to the meridian, and bring the XII to the meridian; then turn the sun's place to the eastern edge of the horizon, and the meridian-will point out the hour of rising; if you bring it to the western edge of the horizon, it will shew the setting.

Thus on the 16th day of March, the sun rose a little past six, and set a little before six.

Note. In the summer the sun rises and sets a little to the northward of the east and west points, but in winter, a little to the southward of them. If, therefore, when the sun's place is brought to the eastern and western edges of the horizon, you look on the inner circle, right against the sun's place, you will see the point of the com­pass upon which the sun rises and sets that day.

PROBLEM X.
To find the length of the day and night at any time of the year.

Only double the time of the sun's rising that day, and it gives the length of the night; doub­le the time of his setting, and it given the length of the day.

This problem shows how long the sun stays with us any day, and how long he is absent from us any night.

Thus on the 26th of May the sun rises about four, and sets about eight; therefore the day is sixteen hours long, and the night eight.

PROBLEM XI.
To find the length of the longest or shortest day, at any place upon the earth.

Rectify the globe for that place, bring the beginning of Cancer to the meridian, bring XII to the meridian, then bring the same de­gree of Cancer to the east part of the horizon, and the meridian will shew the time of the sun's rising.

If the same degree be brought to the western side, the meridian will shew the set­ting, which doubled, (as in the last problem) [Page 148] will give the length of the longest day and short­est night.

If we bring the beginning of Capricorn to the meridian, and proceed in all respects as be­fore, we shall have the length of the longest night and shortest day.

Thus in the Great Mogul's dominions, the longest day is fourteen hours, and the shortest night ten hours. The shortest day is ten hours, and the longest night fourteen hours.

At Petersburgh, the seat of the Empress of Russia, the longest day is about 19½ hours, and the shortest night 4½ hours; the shortest day 4½ hours, and longest night 19½ hours.

Note. In all places near the equator, the sun rises and sets at six the year round. From thence to the polar circles, the days in­crease as the latitude increases; so that at those circles themselves, the longest day is 24 hours. and the longest night just the same. From the polar circles to the poles, the days continue to lengthen into weeks and months; so that at the very pole, the sun shines for six months toge­ther in summer, and is absent from it six months in winter.

Note. That when it is summer with the northern inhabitants, it is winter with the sou­thern, and the contrary; and every part of the world partakes of an equal share of light and darkness.

PROBLEM XII.
To find all those inhabitants to whom the sun is this moment rising or setting, in their meridian or midnight.

Find the sun's place in the ecliptic, and raise the pole as much above the horizon as the sun (that day) declines from the equator; then bring the place where the sun is vertical at that hour to the brass meridian; so will it then be in the zenith or center of the horizon. Now see what countries lie on the western edge of the horizon, for in them the sun is rising; to those on the eastern side he is setting; to those under the upper part of the meridian it is noon day; and to those under the lower part of it, it is midnight.

Thus on the 25th of April, at six o'clock in the evening, at Worcester,

The sun is rising at New Zealand; and to those who are sailing in the middle of the Great South Sea.

The sun is setting at Sweden, Hungary, Italy, Tunis, in the middle of Negroland and Guinea.

In the meridian (or noon) at the middle of Mexico, Bay of Honduras, middle of Florida, Canada, &c.

[Page 150] Midnight at the middle of Tartary, Bengal, India, and the seas near the Sunda isles.

PROBLEM XIII.
To find the beginning and end of twilight.

The twilight is that faint light which opens the morning by little and little in the east, be­fore the sun rises; and gradually shuts in the evening in the west, after the sun is set. It arises from the sun's illuminating the upper part of the atmosphere, and begins always when he approaches within eighteen degrees of the eastern part of the horizon, and ends when he descends eighteen degrees below the western; when dark night commences, and continues till day breaks again.

To find the beginning of twilight, rectify the globe; turn the degree of the ecliptic, which is opposite to the sun's place, till it is ele­vated eighteen degrees in the quadrant of alti­tude above the horizon on the west, so will the index point the hour twilight begins.

This short specimen of problems by the old globes, it is presumed, will be sufficient to ena­ble the pupil to solve any other.

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, in­clined in an angle of 23½ degrees to the broad pa­per 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 re­presents the equinoctial, will not be altered by such a revolution of the intersecting or equi­noctial points. This motion is called the pre­cession 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 [Page 160] equinox. It is a very slow motion, for the equi­noctial points take up 72 years to move one degree, and therefore they are 25,920 years in describing 360 degrees, or completing a revo­lution.

This motion of the poles is easily repre­sented 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 there­to, 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 repre­sent 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.^*

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 dis­tinguished into seven different magnitudes.

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 de­grees, 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 gra­duated edge of the strong brazen meridian, and set the hour index to the most elevated twelve.

[Page 164] 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 be­tween a degree of the equinoctial at the hori­zon 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 lat­itude, 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 in­dex 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 Ca­pricorn; 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 long­est 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 coin­cides 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-set­ting 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.

[Page 169] The longest day at all places in north lati­tude, is when the sun is in the first point of Can­cer. And,

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

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 ele­vated, on turning the globe from east to west, until the artificial sun coincides with the wes­tern 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 short­est 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 cli­mate. This holds good for every climate be­tween 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 [Page 170] 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 po­lar 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 be­gins.

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 ho­rizon, 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 [Page 171] 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 ho­rizon; 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.

PROBLEM XI.
To illustrate, by the globe, so much of the equation of time as is in consequence of the sun's ap­parent 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 cal­led 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 be­fore, sometimes behind the sun.

[Page 175] A good and well-regulated clock goes uni­formly 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 follow­ing. The time measured by a clock is called equal or true time, that measured by the sun­dial apparent time.

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 decli­nation; 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 me­ridian; 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 cata­logue of the fixed stars.

PROBLEM XIII.
To find the latitude and longitude of a given star.

Bring the pole of the ecliptic to the meri­dian, over which fix the quadrant of altitude, and, holding the globe very steady, move the quadrant to lie over the given star, and the de­gree on the quadrant cut by the star, is it's la­titude; 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 mo­tion, whether any stars vanish, or new ones ap­pear.

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▪ [Page 178] 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 nor­thern part, you will observe the star Aldebaran, or the bull's eye.

PROBLEM XV.
To find at what hour any known star passes the me­ridian, at any given day.

Find the sun's place for that day in the ecliptic, and bring it to the strong brass meri­dian, set the horary index to XII o'clock, then turn the globe till the star comes to the meri­dian, and the index will mark the time. Thus on the 15th of August, Lyra comes to the me­ridian 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 calen­dar of the broad paper circle.

By knowing whether the hour be in the morning or afternoon, it will be easy to per­ceive 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 after­noon.

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 [Page 180] 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 vi­sible 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, [Page 181] 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 Arc­turus, 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 beauti­ful 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 [Page 182] about 42° high, is the constellation Cassiopeia: and almost in the north, near the horizon, is the constellation Cygnus.^*

PROBLEM XVIII.
To trace the circles of the sphere in the starry fir­mament.

I shall solve this problem for the time of the autumnal equinox; because that intersec­tion 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 meri­dians, 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 corres­pond 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 [Page 183] of Hercules, the shoulders of Serpentarius, and Sobieski's shield; it passes a little below the feet of Antinous, and through those of Capri­corn, through the Sculptor's frame, Eridanus, the star Rigel in Orion's foot, the head of Mo­noceros, 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 a­bove the horizon, if we begin from the western side, presents to our view Capricornus, Aqua­rius, Pisces, Aries, Taurus, Gemini, and a part of the constellation Cancer.

[Page 184] The solstitia colure, from the western side, passes through Cerberus, and the hand of Her­cules, thence by the western side of the constel­lation 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, ad­vancing 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 Cameloparda­lus, 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 con­stellation 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 Pol­lux, and so through a part of the constellation [Page 185] 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 Aqua­rius, 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 depres­sed under the horizon.

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

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

The largest parallel of latitude on the terres­trial globe, as well as the largest circle of decli­nation 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 ter­restrial, [Page 186] and the largest parallel of declination on the celestial globe, which is entirely hid be­low the horizon of any place, was by the an­cients 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 in­tersection 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 de­grees, 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 con­sidered as both an arctic and antarctic circle, [Page 187] 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 arc­tic, 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 ze­nith 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 meri­dian, between the north and south poles of the earth.

Many authors have mistaken these mutable circles, and have given their names to the im­mutable 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 appa­rition, 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 perpe­tual apparition is that parallel of declination which is equal to the complement of the given latitude northward; and for the perpetual occul­tation, it is the same parallel southward; that is to say, in other words, all those stars, whose declinations exceed the co-latitude, will al­ways be visible, or above the horizon; and all those in the opposite hemisphere, whose declina­tion 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.

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 am­plitude.

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 ampli­tude at rising.

If the same point be brought to the western side of the horizon, the arch of the horizon in­tercepted 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 am­plitude.

The amplitude of the sun at rising and set­ting increases with the latitude of the place: and in very high northern latitudes, the sun scarce sets before he rises again. Homer had [Page 190] heard something of this, though it is not true of the Laestrygones, to whom he applies it.

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 rec­tify 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 [Page 191] 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 co­latitude 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.

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 qua­drant, and the index will shew the hour.

If the latitude and declination are of dif­ferent 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 [Page 194] 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 ze­nith, bring the sun's place for the day to the me­ridian, 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 wes­tern amplitude, oblique descension, and time of setting, will be found. 4. The time of ris­ing, 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 [Page 195] London the star Sirius; by working the pro­blem, you will find

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

PROBLEM XXIV.
The latitude, the altitude of the SUN by day, or of aSTAR by night, being given, to find the hour of the day, and the sun's or star's azi­muth.

Rectify the globes for the latitude, the ze­nith, 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 azi­muth 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 ze­nith, 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 dis­tance; 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 al­titude 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 azi­muth 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 azi­muth 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 depres­sion and the altitude required.

Thus if the time given be the 1st of No­vember, 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 ze­nith, and the sun's place; then put the qua­drant 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 [Page 198] 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 azi­muth, 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 [Page 199] 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 azi­muth 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 ze­nith, and the sun's place; then turn the globe, [Page 200] 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 com­mon 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 lati­tude of the place, especially at sea, how this may be found, in a rude manner at least, hav­ing [Page 201] 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 hori­zon, 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 me­ridian up or down in the notches, till the other star is brought to the east or west part of the ho­rizon, and that elevation of the pole will be the the latitude of the place sought.

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 lati­tude 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 Scor­pio, 29 deg. opposite to which, on the horizon, will be found November the 21st. The Lu­cida Pleiadum, therefore, sets cosmically No­vember the 21st.

In the same manner, in the latitude of Lon­don, Sirius will be found to rise cosmically August the 10th, and to set cosmically No­vember the 10th.

It is of the cosmical setting of the Pleiades, [Page 204] that Virgil is to be understood in this line,

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

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 lati­tude 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. There­fore Arcturus rises at Athens achronically April the 2d.

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

[Page 205] If the globe continues rectified to the lati­tude 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 achroni­cally 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.

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 lati­tude 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 qua­drant of altitude; that will be found to be Ca­pricorn, 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 [Page 207] 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."^*

PROBLEM XXXVII.
The latitude of the place being given, to find the time of the year when a given star sets heli­acally.

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 [Page 208] 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 oppo­site to which, in the ecliptic, is 7 deg. of Ge­mini; 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.

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 lati­tude in an ephemeris, and set the graduated edge of the moveable meridian to the given longitude in the ecliptic, and counting so ma­ny degrees amongst the parallels in the zodiac, either above or below the ecliptic, as her lati­tude is north or south; and set the center of the [Page 213] 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 hori­zon, 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 lati­tude: then bring the place of the sun to the western semicircle of the horizon, and observe [Page 214] 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 obser­vation 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 la­titude 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 differ­ence in the process, no repetition can be ne­cessary.

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.

[Page 221] But if you are provided with an ephemeris,^* 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 de­grees 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 ar­tificial 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 lat­itude 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 al­titudes 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 e­quinox; 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 there­fore be 57 deg. difference in the meridian alti­tude 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 pheno­menon 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 enter­tainment 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 ob­served 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 im­mediate 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 con­spicuous, 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 circum­stances or seasons render it more or less service­able.^*

About the equator, where there is no variety of seasons, moon-light is not necessary for ga­thering in the produce of the ground; and [Page 224] 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 set­ting, from the first to the third quarter.

But this observation is still further con­firmed, when we consider that this appearance is only peculiar with respect to the full moon, from which only the farmer can derive any ad­vantage; 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 diffe­rent 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 an­gles.

The moon's motion is so nearly in the [Page 225] ecliptic, that we may consider her at present as moving in it.

The different parts of the ecliptic, on ac­count 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 hori­zon; but when the sun is in the autumnal equi­nox, 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 in­clination of the ecliptic to the horizon, so a greater or less degree of motion of the globe [Page 226] 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 pro­portion; 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 au­tumnal 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 sun­set 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 [Page 227] 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 eclip­tic 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 pro­blem, on the globe.

[Page 228] 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 ephe­meris, to which set the artificial moon,^* 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 [Page 229] which the moon was, and the degrees on the horizon contained between the strong brass me­ridian 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 zodi­ac; put the quadrant of altitude to the azimuth before found, and turn the globe till the artifi­cial 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.^*

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 cen­tripetal force, every where decreasing as the [Page 230] 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 mo­tions 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 ap­pearance 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 ren­dered [Page 231] 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-men­tioned 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 pro­blem.

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 al­titude 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 [Page 322] 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 la­titude, 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 morn­ing; then bring the quadrant of altitude to in­tersect the horizon in 71 deg. the given azi­muth 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 [Page 233] it's longitude sought. Then bring the come­tary patch to the brazen meridian, and it's de­clination 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 ob­servation 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 set­ting.

N. B. From hence it appears, the comet rose soon enough that morning to have been ob­served at London, had the heavens been clear, and the astronomers been before-hand appriz­ed 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 [Page 235] two fixed stars marked μ and ν in the constel­lation 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 deter­mined 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 be­tween 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 re­quired. And here it may be observed, that [Page 236] it's apparent path lay through the following southern constellations, viz. the tail of Capri­corn, 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-men­tioned; after this, it passed on to Sextans Ura­niae, 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 be­ginning of it's appearance, and the other to­wards 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 coin­cide 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 ele­vated 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.

[Page 238] After this manner may the problems relat­ing to any other comets be performed; and thus the paths of the several comets, which have hitherto been observed, may be severally deli­neated on the celestial globe, and their various phenomena in different latitudes be thereby shewn.

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