Chapter IV: Part 4

In the preceding problem, the globe being rectified according to the Sun’s declination, the upper parts of the parallels of latitude, represented the _Diurnal Arches_, or the length of the days all over the world, at that particular time: Here we are to rectify the globe according to the latitude of the place, and then the upper parts of the parallels of declination are the diurnal arches; and the length of the days at all times of the year, may be here determined by finding the number of hours contained between the two extreme meridians, which cut any parallel of declination in the Eastern and Western points of the horizon; after the same manner, as before we found the length of the day in the several latitudes at a particular time of the year.

1. Let the place proposed be under the equinoctial, and let the globe be accordingly rectified for 00 degrees of latitude, which is called a direct position of the sphere. Here all the parallels of latitude, which in this case we will call the parallels of declination, are cut by the horizon into two equal parts; and consequently those who live under the equinoctial, have the days and nights of the same length at all times of the year; and also in this part of the Earth, all the _Stars_ rise and set, and their continuance above the horizon, is equal to their stay below it, _viz._ 12 hours.

If from this position we gradually move the globe according to the several alterations of latitudes, which we will suppose to be Northerly; the lengths of the _Diurnal Arches_ will continually increase, until we come to a parallel of declination, as far distant from the equinoctial, as the place itself is from the Pole. This parallel will just touch the horizon, and all the heavenly bodies that are betwixt it and the Pole never descend below the horizon. In the mean time, while we are moving the globe, the lengths of the diurnal arches of the Southern parallels of declination, continually diminish in the same proportion that the Northern ones increased; until we come to that parallel of declination which is so far distant from the equinoctial Southerly, as the place itself is from the North Pole. The upper part of this _Parallel_ just touches the horizon, and all the Stars that are betwixt it and the South Pole never appear above the horizon. And all the nocturnal arches of the Southern parallels of declination, are exactly of the same length with the diurnal arches of the correspondent parallels of North declination.

2. Let us take a view of the globe when it is rectified for the latitude of _London_, or 51½ degrees North. When the Sun is in the tropic of ♋, the day is about 16½ hours; as he recedes from this tropic, the days proportionably shorten, until, he arrives into ♑, and then the days are at the shortest, being now of the same length with the night, when the Sun was in ♋, _viz._ 7½ hours. The lower part of that parallel of declination, which is 38½ degrees from the equinoctial Northerly, just touches the horizon; and the Stars that are betwixt this parallel and the North Pole, never set to us at _London_. In like manner the upper part of the Southern parallel of 38½ degrees just touches the horizon, and the Stars that lie betwixt this parallel and the Southern Pole, are never visible in this latitude.

Again, let us rectify the globe for the latitude of the _Arctic Circle_, we shall then find, that when the Sun is in ♋, he touches the horizon on that day without setting, being 24 hours compleat above the horizon; and when he is in _Capricorn_, he once appears in the horizon, but does not rise in the space of 24 hours: When he is in any other point of the ecliptic, the days are longer or shorter, according to his distance from the tropics. All the Stars that lie between the tropic of _Cancer_, and the North Pole, never set in this latitude; and those that are between the tropic of _Capricorn_, and the South Pole, are always hid below the horizon.

If we elevate the globe still higher, the circle of _perpetual Apparition_ will be nearer the equator, as will that of _perpetual Occultation_ on the other side. For example, Let us rectify the globe for the latitude of 80 degrees North: when the Sun’s declination is 10 degrees North; he begins to turn above the horizon without setting; and all the while he is making his progress from this point to the tropic of ♋, and back again, he never sets. After the same manner, when his declination is 10 degrees South, he is just seen at noon in the horizon; and all the while he is going Southward, and back again, he disappears, being hid just so long as before, at the opposite time of the year he appeared visible.

Let us now bring the North Pole into the Zenith, then will the equinoctial coincide with the horizon; and consequently all the Northern parallels are above the horizon, and all the Southern ones below it. Here is but one day and one night throughout the year, it being day all the while the Sun is to the Northward of the equinoctial, and night for the other half year. All the Stars that have North declination, always appear above the horizon, and at the same height; and all those that are on the other side, are never seen.

What has been here said of rectifying the globe to North latitude, holds for the same latitude South; only that before the longest days were, when the Sun was in ♋, the same happening now when the Sun is in ♑; and so of the rest of the parallels, the seasons being directly opposite to those who live in different hemispheres.

I shall again explain some things delivered above in general terms, by particular problems.

But from what has been already said, we may first make the following observations:

1. _All places of the Earth do equally enjoy the benefit of the Sun, in respect of time, and are equally deprived of it, the Days at one time of the Year, being exactly equal to the Nights at the opposite season._

2. _In all places of the Earth, save exactly under the Poles, the Days and Nights are of equal length_ (viz. _12 hours each) when the Sun is in the equinoctial._

3. _Those who live under the equinoctial, have the days and nights of equal lengths at all times of the year._

4. _In all places between the equinoctial and the Poles, the days and nights are never equal, but when the Sun is in the equinoctial points_ ♈ _and_ ♎.

5. _The nearer any place is to the equator, the less is the difference between the length of the artificial days and nights in the said place; and the more remote the greater._

6. _To all the inhabitants lying under the same parallel of latitudes the days and nights are of equal lengths, and that at all times of the year._

7. _The Sun is vertical twice a year to all places between the tropics; to those under the tropics, once a year; but never any where else._

8. _In all places between the Polar Circles, and the Poles, the Sun appears some number of days without setting; and at the opposite time of the year he is for the same length of time without rising; and the nearer unto, or further remote from the Pole, those places are, the longer or shorter is the Sun’s continued presence or absence from the Pole._

9. _In all places lying exactly under the Polar Circles, the Sun, when he is in the nearest tropic, appears 24 hours without setting; and when he is in the contrary tropic, he is for the same length of time, without rising; but at all other times of the year, he rises and sets there, as in other places._

10. _In all places lying in the (Northern/Southern) hemisphere, the longest day and shortest night, is when the Sun is in the (Northern/Southern) tropic, and on the contrary._

PROB. XVIII. _The Latitude of any place, not
exceeding 66½ degrees, and the day of the Month
being given; to find the time of Sun-rising and
setting, and the length of the Day and Night._

Having rectified the globe according to the latitude, bring the Sun’s place to the meridian, and put the hour index to 12 at noon; then bring the Sun’s place the Eastern part of the horizon, and the index will shew the time when the Sun rises. Again, turn the globe until the Sun’s place be brought to the Western side of the horizon, and the index will shew the time of Sun-setting.

The hour of Sun-setting doubled, gives the length of the day; and the hour of Sun-rising doubled, gives the length of the night.

Let it be required to find when the Sun rises and sets at _London_ on the 20th of _April_. Rectify the globe for the latitude of _London_, and having found the Sun’s place corresponding to _May_ the 1st, _viz._ ♉ 10¾ degrees, bring ♉ to 10¾ degrees to the meridian, and set the index to 12 at noon; then turn the globe about ’till ♉ 10¾ degrees be brought to the Eastern part of the horizon, and you’ll find the index point 4¾ hours, this being doubled, gives the length of the night 9½ hours. Again, bring the Sun’s place to the Western part of the horizon, and the index will point 7¼ hours, which is the time of Sun-setting; this being doubled, gives the length of the day 14½ hours.

PROB. XIX. _To find the length of the longest
and shortest Day and Night in any given place, not
exceeding 66½ degrees of Latitude._

_Note_, The longest day at all places on the (North/South) side of the equator, is when the Sun is in the first point of (_Cancer_/_Capricorn_) Wherefore having rectified the globe for the latitude, find the time of Sun-rising and setting, and thence the length of the day and night, as in the last problem, according to the place of the Sun: Or, having rectified the globe for the latitude, bring the solstitial point of that hemisphere, to the East part of the horizon, and set the index to 12 at noon; then turning the globe about ’till the said solstitial point touches the Western side of the horizon, the number of hours from noon to the place where the index points (being counted according to the motion of the index) is the length of the longest day; the complement whereof to 24 hours, is the length of the shortest night, and the reverse gives the shortest day and the longest night.

_Longest Day._ _Shor. N._
_Deg._ _Hours._ _Hours._
{ 45 15½ 8½
Thus in Lat. { 51½ 16½ 7½
{ 60 18½ 5½

If from the length of the longest day, you subtract 12 hours, the number of half hours remaining, will be the _Climate_: Thus that place where the longest day is 16½ hours, lies in the 9th _Climate_. And by the reverse, having the _Climate_, you have thereby the length of the longest day.

PROB. XX. _To find in what Latitude the
longest Day is, of any given length, less than 24
hours._

Bring the solstitial point to the meridian, and set the index to 12 at noon; then turn the globe Westward, ’till the index points at half the number of hours given; which being done, keep the globe from turning round its axis, and slide the meridian up or down in the notches, ’till the solstitial point comes to the horizon, then that elevation of the Pole will be the latitude.

If the hours given be 16, the latitude is 49 degrees; if 20 hours, the latitude is 63¼ degrees.

PROB. XXI. _A place being given in one of the_
Frigid Zones (_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 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.

PROB. XXII. _To find in what Latitude the
longest Day is, of any given length less than_ 182
_Natural Days._

Find a point in the ecliptic half so many degrees distant from the solstitial point, as there are days given, and bring that point to the meridian; then keep the globe from turning round its axis, and move the meridian up or down until the aforesaid point of the ecliptic comes to the horizon; that elevation of the Pole will be the latitude required.

If the days given were 78, the latitude is 71½ degrees.

This method is not accurate, because the degrees in the ecliptic do not correspond to natural days; and also because the Sun does not always move in the ecliptic at the same rate; however, such problems as these may serve for amusements.

PROB. XXIII. _The day of the Month being
given, to find when the Morning and Evening_
Twilight _begins and ends, in any place upon the
Globe._

In the foregoing problem, by the length of the day, we mean the time from Sun-rising to Sun-set; and the night we reckoned from Sun-set, ’till he rose next morning. But it is found by experience, that _Total Darkness_ does not commence in the evening, ’till the Sun has got 18 degrees below the horizon; and when he comes within the same distance of the horizon next morning, we have the first _Dawn of Day_. This faint light which we have in the morning and evening, before and after the Sun’s rising and setting, is what we call the _Twilight_. [4] Having rectified the globe for the latitude, the zenith, and the Sun’s place, turn the globe and the quadrant of altitude until the Sun’s place cuts 18 degrees below the horizon (if the quadrant reaches so far) then the index upon the hour circle will shew the beginning or ending of twilight after the same manner as before we found the time of the Sun-rising and setting, in _Prob. 18_. But by reason of the thickness of the wooden horizon, we can’t conveniently see, or compute when the Sun’s place is brought to the point aforesaid. Wherefore the globe being rectified as above directed, turn the globe, and also the quadrant of altitude, Westward, until that point in the ecliptic, which is opposite to the Sun’s place, cuts the quadrant in the 18th degree above the horizon; then the hour index will shew the time when day breaks in the morning. And if you turn the globe and the quadrant of altitude, until the point opposite to the Sun’s place cuts the quadrant in the Eastern hemisphere, the hour hand will shew when twilight ends in the evening. Or, having found the time from midnight when the morning twilight begins, if you reckon so many hours before midnight, it will give the time when the evening twilight ends. Having found the time when twilight begins in the morning, find the time of Sun-rising, by _Prob. 18_, and the difference will be the duration of twilight.

Thus at _London_ on the 12th of _May_ twilight begins at three quarters past one o’clock: The Sun rises at about half an hour past four: Whence the duration of twilight now is 2¾ hours, both in the morning and evening. On the 12th of _November_, the twilight begins at half an hour past six, being somewhat above an hour before Sun-rising.

PROB. XXIV. _To find the time when total
Darkness ceases, or when the Twilight continues
from Sun-setting to Sun-setting, in any given
place._

Let the place be in the Northern hemisphere; then if the complement of the latitude be greater than (the depression) 18 degrees, subtract 18 degrees from it, and the remainder will be the Sun’s declination North, when total darkness ceases. But if the complement of the latitude is less than 18 degrees, their difference will be the Sun’s declination South, when the twilight begins to continue all night. If the latitude is South, the only difference will be, that the Sun’s declination will be on the contrary side.

Thus at _London_, when the Sun’s declination North is greater than 20½ degrees, there is no total darkness, but constant twilight, which happens from the 26th of _May_ to the 18th of _July_, being near two months. Under the North Pole the twilight ceases, when the Sun’s declination is greater than 18 degrees South, which is from the 13th of _November_, ’till the 29th of _January_: So that notwithstanding the Sun is absent in this part of the world for half a year together, yet total darkness does not continue above 11 weeks; and besides, the _Moon_ is above the horizon for a whole fortnight of every month throughout the year.

PROB. XXV. _The day of the Month be given; to
find those places of the Frigid Zones, where the
Sun begins to shine continually without setting;
and also those places where he begins to be totally
absent._

Bring the Sun’s place to the meridian, and mark the number of degrees contained betwixt that point and the equator; then count the same number of degrees from the nearest Pole (_viz._ the North Pole, if the Sun’s declination is Northerly, otherwise the South Pole) towards the equator, and note that point upon the meridian; then turn the globe about, and all the places which pass under the said point, are those where the Sun begins to shine constantly, without setting on the given day. If you lay the same distance from the opposite Pole towards the equator, and turn the globe about, all the places which pass under that point, will be those where the longest night begins.

_The Latitude of the place being given, to find the
hour of the day when the Sun shines._

_If it be in the summer_, elevate the Pole according to the latitude, and set the meridian due North and South; then the shadow of the axis will cut the hour on the Dial plate: For the globe being rectified in this manner, the hour circle is a true _Equinoctial Dial_; the axis of the globe being the _Gnomon_. This holds true in _Theory_, but it might not be very accurate in practice, because of the difficulty in placing the horizon of the globe truly horizontal, and its meridian due North and South.

If it be in the winter half year, elevate the South Pole according to the latitude North, and let the North part of the horizon be in the South part of the meridian; then the shade of the axis will show the hour of the day as before: But this cannot be so conveniently performed, tho’ the reason is the same as in the former case.

_To find the Sun’s altitude, when it shines, by the Globe._

Having set the frame of the globe truly horizontal or level, turn the North Pole towards the Sun, and move the meridian up or down in the notches, until the axis casts no shadow; then the arch of the meridian, contained betwixt the Pole and the horizon, is the Sun’s altitude.

_Note_, The best way to find the Sun’s altitude, is by a little quadrant graduated into degrees, and having sights and a plummet to it: Thus, hold the quadrant in your hand, so as the rays of the Sun may pass through both the sights, the plummet then hanging freely by the side of the instrument, will cut in the limb the altitude required. These quadrants are to be had at the instrument-makers, with lines drawn upon them, for finding the hour of the day, and the azimuth; with several other pretty conclusions, very entertaining for beginners.

_The Latitude and the Day of the Month being given, to
find the hour of the day when the Sun shines._

Having placed the wooden frame upon a level, and the meridian due North and South, rectify the globe for the latitude, and fix a needle perpendicularly over the Sun’s place: The Sun’s place being brought to the meridian, set the hour index at 12 at noon, then turn the globe about until the needle points exactly to the Sun, and casts no shadow, and then the index will shew the hour of the day.

PROB. XXVI. _The Latitude, the Sun’s Place,
and his Altitude, being given; to find the hour of
the Day, and the Sun’s Azimuth from the Meridian._

Having rectified the globe for the latitude, the zenith, and the Sun’s place, turn the globe and the quadrant of altitude, so that the Sun’s place may cut the given degree of altitude: then the index will show the hour, and the quadrant will cut the azimuth in the horizon. Thus, if at _London_, on the 21st of _August_, the Suns altitude, be 36 degrees in the forenoon, the hour of the day will be IX, and the Sun’s azimuth about 58 degrees from the South part of the meridian.

_The Sun’s Azimuth being given, to place the Meridian
of the Globe due North and South, or to find a
Meridian Line when the Sun shines._

Let the Sun’s azimuth be 30 degrees South-Easterly, set the horizon of the globe upon a level, and bring the North Pole into the zenith; then turn the horizon about until the shade of the axis cuts as many hours as is equivalent to the azimuth (allowing 15 degrees to an hour) in the North-West part of the hour circle, _viz._ X at night, which being done, the meridian of the globe stands in the true meridian of the place. The globe standing in this position, if you hang two plummets at the North and South points of the wooden horizon, and draw a line betwixt them, you will have a meridian line; which if it be on a fixed plane (as a floor or window) it will be a guide for placing the globe due North and South, at any other time.

PROB. XXVII. _The Latitude, Hour of the Day,
and the Sun’s place being given, to find the Sun’s
Altitude and Azimuth._

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 degrees, 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 61½ degrees; and at 10 o’clock in the morning, the Sun’s altitude will be 52 degrees, and his azimuth about 50 degrees from the South part of the meridian.

PROB. XXVIII. _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 _December_, at 10 o’clock at night, the depression and azimuth will be the same as was found in the last problem.

PROB. XXIX. _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 meet the edge of the quadrant, then the said edge will shew the altitude, and the index point to the hour. Thus, _May_ the 21st at _London_ when the Sun is due East, his altitude will be about 24 degrees, 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 about 2¾ 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.

PROB. XXX. _The Latitude, the Sun’s Altitude,
and his Azimuth being given; to find his Place in
the Ecliptic and the Hour._

Rectify the globe for the latitude and zenith, and set the edge of the quadrant to the given azimuth; then turning the globe about, that point of the ecliptic which cuts the altitude, will be the Sun’s place. Keep the quadrant of the altitude in the same position, and having brought the Sun’s place to the meridian, and the hour index to 12 at noon, turn the globe about ’till the Sun’s place cuts the quadrant of altitude, and then the index will point the hour of the day.

PROB. XXXI. _The Declination and Meridian
Altitude of the Sun, or of any Star being given; to
find the Latitude of the Place._

Mark the point of declination upon the meridian, according as it is either North or South from the equator; then slide the meridian up or down in the notches, ’till the point of declination be so far distant from the horizon, as is the given meridian altitude; that elevation of the Pole will be the latitude.

Thus, if the Sun’s, or any Star’s meridian altitude be 50 degrees, and its declination 11½ degrees North, the latitude will be 51½ degrees North.

PROB. XXXII. _The Day and Hour of a Lunar Eclipse being known; to find all those Places upon the Globe where the same will be visible._ [5] Find where the Sun is vertical at the given hour, and bring that point to the zenith; then the Eclipse will be visible in all those places that are under the horizon; Or, if you bring the Antipodes to the place where the Sun is vertical, into the zenith, you will have the places where the Eclipse will be visible above the horizon.

_Note_, Because _Lunar_ eclipses continue sometimes for a long while together, they may be seen in more places than one hemisphere of the Earth; for by the Earth’s motion round its axis, during the time of the eclipse, the Moon will rise in several places after the eclipse began.

_Note_, When an eclipse of the Sun is central, if you bring the place where the Sun is vertical at that time, into the zenith, some part of the eclipse will be visible in most places within the upper hemisphere; but by reason of the short duration of Solar eclipses, and the latitude which the Moon commonly has at that time (tho’ but small) there is no certainty in determining the places where those eclipses will be visible by the globe; but recourse must be had to calculations.

PROB. XXXIII. _The Day of the Month, and Hour
of the Day, according to our way of reckoning
in_ England, _being given; to find thereby the_
Babylonic, Italic, _and the_ Jewish, _or Judaical
Hour._

1. To find the _Babylonic Hour_ (which is the number of hours from Sun-rising.) Having found the time of Sun-rising in the given place, the difference betwixt this and the hour given, is the _Babylonic Hour_.

2. To find the _Italic Hour_ (which is the number of hours from Sun-setting.) Subtract the hour of Sun-setting from the given hour, and the remainder will be the _Italic Hour_ required.

3. To find the _Jewish Hour_ (which is ¹/₁₂ part of an _Artificial Day_.) Find how many hours the day consists of; then say, as the number of hours the day consists of is to 12 hours, so is the hour since Sun-rising to the _Judaical_ hour required.

Thus, if the Sun rises at 4 o’clock (consequently sets at 8) and the hour given be 5 in the evening, the _Babylonish_ hour will be the 13th, the _Italic_ the 21st and the _Jewish_ hour will be nine and three quarters.

The converse being given, the hour of the day, according to our way of reckoning in _England_, may be easily found.

The following Problems are peculiar to the _Celestial Globe_.

PROB. XXXIV. _To find the Right Ascension and
Declination of the Sun, or any Fixed Star._

Bring the Sun’s place in the ecliptic to the meridian; then that degree of the equator, which is cut by the meridian, will be the _Sun’s Right Ascension_; and that degree of the meridian, which is exactly over the Sun’s place, is the _Sun’s Declination_.

After the same manner, bring the place of any Fixed Star to the meridian, and you will find its Right Ascension in the equinoctial, and Declination of the meridian.

Thus, the right ascension and declination is found, after the same manner as the longitude and latitude of a place upon the _Terrestrial Globe_.

_Note_, The right ascension and declination of the Sun vary every day; but the right ascension, _&c._ of the Fixed Stars is the same throughout the year[6].

The Sun’s Right Ascension. Declin.
_Deg._ _Deg._
{ _January_ 31 314 17⅓ S.
{ _April_ 5 14¼ 6 N.
Thus on { _July_ 21 120¼ 20½ N.
{ _November_ 26 242¼ 21 S.

R. Asc. Dcl.
_Deg._ _Deg._
_Aldebaran_ 65 16 N.
_Spica Virginis_ 197¾ 9¾ S.
_Capella_ 74 45⅔ N.
_Syrius, or the Dog-Star_ 98¼ 16⅓ S.

_Note_, The declination of the Sun may be found after the same manner by the _Terrestrial Globe_, and also his right ascension, when the equinoctial is numbered into 360 degrees, commencing at the equinoctial point ♈: But as the equinoctial is not always numbered so, and this being properly a Problem in _Astronomy_, we choose rather to place it here.

By the converse of this problem, having the right ascension and declination of any point given, that point itself may be easily found upon the globe.

PROB. XXXV. _To find the Longitude and
Latitude of a given Star._

Having brought the solstitial colure to the meridian, fix the quadrant of altitude over the proper Pole of the ecliptic, whether it be North or South; then turn the quadrant over the given Star, and the arch contained betwixt the Star and the ecliptic, will be the latitude, and the degree cut on the ecliptic will be the Star’s longitude.

Thus the latitude of _Arcturus_ will be found to be 31 degrees North, and the longitude 200 degrees from ♈, or 20 degrees from ♎: The latitude of _Fomalhaut_ in the Southern Fish, 21 degrees South, and longitude 299½ degrees, or ♑ 29½ degrees. By the converse of this method, having the latitude and longitude of a Star given, it will be easy to find the Star upon the globe.

The distance betwixt two Stars, or the number of degrees contained betwixt them, may be found by laying the quadrant of altitude over each of them, and counting the number of degrees intercepted; after the same manner as we found the distance betwixt two places on the _Terrestrial_ Globe, in _Prob._ VII.

PROB. XXXVI. _The Latitude of the Place, the
Day of the Month, and the Hour being given; to
find what Stars are then rising or setting, what
Stars are culminating, or on the meridian, and
the Altitude and Azimuth of any Star above the
Horizon; and also how to distinguish the Stars in
the Heavens one from the other, and to know them by
their proper Names._

Having rectified the globe for the latitude, the zenith, and the Sun’s place, turn the globe about until the index points to the given hour, the globe being kept in this position.

All those Stars that are in (Eastern/Western)
side of the horizon, are then (Rising/Setting).

All those Stars that are under the meridian, are then culminating. And if the quadrant of altitude be laid over the center of any particular Star, it will show that Star’s altitude at that time; and where it cuts the horizon, will be the Star’s azimuth from the North or South part of the meridian.

The globe being kept in the same elevation, and from turning round its axis, move the wooden frame about until the North and South points of the horizon lie exactly in the meridian; then right lines imagined to pass from the center thro’ each Star upon the surface of the globe, will point out the real Star in the heavens, which those on the globe are made to represent. And if you are by the side of some wall whose bearing you know, lay the quadrant of altitude to that bearing in the horizon, and it will cut all those Stars which at that very time are to be seen in the same direction, or close by the side of the said wall. Thus knowing some of the remarkable Stars in any part of the heavens, the neighbouring Stars may be distinguished by observing their situations with respect to those that are already known, and comparing them with the Stars drawn upon the globe.

Thus, if you turn your face towards the North, you will find the North Pole of the globe points to the _Pole-Star_; then you may observe two Stars somewhat less bright than the Pole-Star, almost in a right line with it, and four more which form a sort of _quadrangle_; these seven Stars make the constellation called the _Little Bear_; the Pole Star being in the tip of the tail. In this neighbourhood you will observe seven bright Stars, which are commonly called _Charles’s Wane_; these are the bright Stars in the _Great Bear_, and form much such another figure with those before-mentioned in the _little Bear_: The two foremost of the square lie almost in a right line with the Pole Star, and are called the _Pointers_, so that knowing the Pointers, you may easily find the Pole-Star. Thus the rest of the Stars in this constellation, and all the Stars in the neighbouring constellations may be easily found, by observing how the unknown Stars lie either in _quadrangles_, _triangles_, or strait lines from those that are already known upon the globe.

After the same manner the globe being rectified, you may distinguish those Stars that are to the Southward of you, and be soon acquainted with all the Stars that are visible in our hemisphere.

_SCHOLIUM._

The globe being rectified to the latitude of any place, if you turn it round its axis, all those Stars that do not go below the horizon during a whole revolution of the globe, never set in that place; and those that do not come above the horizon never rise.

PROB. XXXVII. _The latitude of the place being
given; to find the Amplitude, Oblique Ascension and
Descension, Ascensional Difference, Semi-diurnal
Arch, and the time of continuance above the
horizon, of any given point in the heavens._

Having rectified the globe for the latitude, and brought the given point to the meridian, set the index to the hour of 12; then turn the globe until the given point be brought to the Eastern side of the horizon, and that degree of the equinoctial which is cut by the horizon at that time, will be the _Oblique Ascension_; and where the given point cuts the horizon, is the _Amplitude Ortive_: If the globe be turned about until the given point be brought to the Western side of the horizon, it will there show the _Amplitude Occasive_; and where the horizon cuts the equinoctial at that time, is the _Oblique Descension_.

The time between the index at either of these two positions, and the hour of 6; or half the difference between the oblique ascension and descension is the _Ascensional Difference_.

If the place be in North latitude and the declination of the given point be (North/South) the ascensional difference reduced into time, and (added to/subtracted from) 6 o’Clock, gives the _Semi-diurnal Arch_; the complement whereof to a semicircle, is the _Semi-nocturnal Arch_. If the place be in South latitude, then the contrary is to be observed with respect to the declination.

The semi-(diurnal/nocturnal) arch being doubled, gives the time of continuance (above/below) the horizon. Or the time of continuance above the horizon, may be found by counting the number of hours contained in the upper part of the horary circle, betwixt the place where the index pointed when the given point was in the Eastern or Western parts of the horizon. If the given point was the Sun’s place, the index pointed the time of his rising and setting, when the said place was in the Eastern and Western parts of the horizon, as in _Prob. 18_. Or the time of Sun-rising may be found by adding or subtracting his ascensional difference, to or from the hour of six, according as the latitude and declination are either contrary or the same way.

Thus, at _London_, on the 31st of _May_, the _Sun’s_

_Amplitude_ is 24 degrees Northerly.
_Oblique Ascension_, 20.
_Oblique Descension_, 58.
_Ascensional Difference_, 19.
_Semi-diurnal Arch_, 109.
His continuance above the horizon, 14½ hours.
Sun rises at three quarters past four.
Sun sets a quarter past seven.

These things for the Sun vary every day; but for a Fixed Star the day of the month need not be given, for they are the same all the year round.

In the latitude of 51½ North, _Syrus_’s
_Amplitude_ is about 28 degrees Southerly.
_Oblique Ascension_, 121.
_Oblique Descension_, 75.
_Ascensional Difference_, 23.
_Semi-diurnal Arch_, 67.
Continuance above the horizon, 9 hours.

PROB. XXXVIII. _The Latitude and the Day of
the Month being given; to find the Hour when any
known Star will be upon the meridian, and also the
time of its rising and setting._

Having rectified the globe for the latitude of the Sun’s place, bring the given Star to the meridian, and also to the East or West side of the horizon, and the index will shew accordingly when the Star _culminates_, or the time of the _rising_ or _setting_.

Thus at _London_, on the 21st of _January_, _Syrius_ will be upon the meridian, at a quarter past ten in the evening; rises at 5¼ hours, and sets at three quarters past two in the morning.

By the converse of this problem, knowing the time when any Star is upon the meridian, you may easily find the Sun’s place. Thus, bring the given Star to the meridian, and set the index to the given hour; then turn the globe ’till the index points to 12 at noon, and the meridian will cut the Sun’s place in the ecliptic. Thus when _Syrius_ comes to the meridian at 10½ hours after noon, the Sun’s place will be ≈ ¼ deg.

PROB. XXXIX. _To find at what time of the Year
a given Star will be upon the Meridian, at a given
Hour of the Night._

Bring the Star to the meridian, and set the index to the given hour, then turn he globe ’till the index points to 12 at noon, and the meridian will cut the ecliptic in the Sun’s place; whence the day of the month may be easily found in the kalendar upon the horizon.

PROB. XL. _The Day of the Month, and the
Azimuth of any known Star being given; to find the
Hour of the Night._

Having rectified the globe for the latitude and the Sun’s place, if the given Star be due North or South, bring it to the meridian, and the index will show the hour of the night. If the Star be in any other direction, fix the quadrant of altitude in the zenith, and set it to the Star’s azimuth in the horizon; then turn the globe about until the quadrant cuts the center of the Star, and the index will shew the hour of the night.

The bearing of any point in the heavens may be found by the following methods.

Having a meridian line drawn in two windows, that are opposite to one another, you may cross it at right angles with another line representing the East and West; from the point of the intersection describe a circle, and divide each quadrant into 90 degrees; then get a smooth board, of about 2 feet long, and ¾ foot broad (more or less, as you judge convenient) and on the back part of it fix another small board crossways, so that it may serve as a foot to support the biggest board upright, when it is set upon a level, or an horizontal plane. The board being thus prepared, set the lower edge of the smooth, or fore side of it, close to the center of the circle, then turn it about to the meridian, or to any azimuth point required (keeping the edge of it always close to the center) and casting your eye along the flat side of it, you will easily perceive what Stars are upon the meridian, or any other bearing that the board is set to.

PROB. XLI. _Two known Stars having the same Azimuth, or
the same Height, being given; to find the Hour of
the Night._

Rectify the globe for the latitude, the zenith, and the Sun’s place.

1. When the two Stars are in the same azimuth, turn the globe, and also the quadrant about, until both Stars coincide with the edge thereof; then will the index shew the hour of the night; and where the quadrant cuts the horizon, is the common azimuth of both Stars.

2. If the two Stars are of the same altitude, move the globe so that the same degree on the quadrant will cut both Stars, then the index will shew the hour.

This problem is useful when the quantity of the azimuth of the two Stars in the first case, or of their altitude in the latter case, is not known.

_If two Stars were given, one on the meridian, and the
other in the East or West part of the horizon; to
find the Latitude._

Bring that Star which was observed on the meridian, to the meridian of the globe, and keep the globe from turning round its axis; then 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 _Latitude_ sought.

PROB. XLII. _The Latitude, Day of the Month,
and the Altitude of any known Star being given; to
find the Hour of the Night._

Rectify the globe for the latitude, zenith, and Sun’s place: Turn the globe, and the quadrant of altitude, backward or forward, ’till the center of that Star meets the quadrant in the degree of altitude given; then the index will point the true hour of the night; and also where the quadrant cuts the horizon, will be the azimuth of the Star at that time.

_If the Latitude, the Sun’s Altitude, and his
Declination (instead of his Place in the Ecliptic)
are given; to find the Hour of the Day and Azimuth._

Rectify the globe for the latitude and zenith, and having brought the _equinoctial colure_ to the meridian, set the index to 12 at noon; which being done, turn the globe and the quadrant, until the given declination in the equinoctial colure, cuts the altitude on the quadrant; then the index will shew the _Hour_ of the day, and the quadrant cut the _Azimuth_ in the horizon.

_If the Altitude of two Stars on the same Azimuth were
given; to find the Latitude of the Place._

Set the quadrant over both Stars at the observed degrees of altitude, and keep it fast upon the globe with your fingers; then slide the meridian up or down in the notches, ’till the quadrant cuts the given azimuth in the horizon; that elevation of the Pole will be the latitude required.

PROB. XLIII. _Having the Latitude of the
place, to find the degree of the Ecliptic, which
rises or sets with a given Star; and from thence
to determine the time of its_ Cosmical _and_
Achronical _rising and setting._

Having rectified the globe for the latitude, bring the given Star to the Eastern side of the horizon, and mark what degree of the ecliptic rises with it: Look for that degree in the wooden horizon, and right against it, in the kalendar, you will find the month and day when the Star _rises Cosmically_. If you bring the Star to the Western side of the horizon, that degree of the ecliptic which rises at that time, will give the day of the month when the said Star _sets Cosmically_. So likewise against the degree which sets with the Star, you will find the day of the month of the _Achronical setting_; and if you bring it to the Eastern part of the horizon, that degree which sets at that time will be the Sun’s place when the Star _rises Achronically_.

Thus, in the latitude of _London_, _Syrius_, or the _Dog-Star_, rises _Cosmically_ the 10th of _August_, and sets _Cosmically_ the 10th of _October_. _Aldebaran_, or the _Bull’s Eye_, rises _Achronically_ on the 22d of _May_, and sets _Achronically_ on the 19th of _December_.

PROB. XLIV. _Having the Latitude of the place,
to find the time when a Star rises and sets_
Heliacally.

Having rectified the globe for the latitude, bring the Star to the Eastern side of the horizon, and turn the quadrant round to the Western side, ’till it cuts the ecliptic in 12 degrees of altitude above the horizon, if the Star be of the first magnitude; then that point of the ecliptic which is cut by the quadrant, is 12 degrees high above the Western part of the horizon, when the Star rises; but at the same time the opposite point in the ecliptic is 12 degrees below the Eastern part of the horizon, which is the depression of a Star of the _first magnitude_, when she _rises Heliacally_; or has got so far from the Sun’s beams, that she may be seen in the morning before Sun-rising. Wherefore look for the said point of the ecliptic on the horizon, and right against it will be the day of the month when the Star _rises Heliacally_. To find the _Heliacal setting_, bring the Star to the West side of the horizon, and turn the quadrant about to the Eastern side, ’till the 12th degree of it above the horizon, cuts the ecliptic; then that degree of the ecliptic which is opposite to this point, is the Sun’s place when the Star _sets Heliacally_.

Thus you will find that _Arcturus_ rises Heliacally the 28th of _September_, and sets Heliacally _December_ the 2d.

PROB. XLV. _To find the place of any Planet
upon the globe; and so by that means, to find its
place in the Heavens: Also to find at what Hour any
Planet will rise or set, or be on the meridian at
any one Day in the Year._

You must first seek in an Ephemeris (_White_’s Ephemeris will do well enough) for the place of the Planet proposed on that day; then mark that point of the ecliptic, either with chalk, or by sticking on a little black patch; and then for that night you may perform any problem, as before, by a Fixed Star.

Let it be required to find the situation of _Jupiter_ among the Fixed Stars in the heavens, and also what time he rises and sets, and comes to the meridian on the 19th of _May_, 1757, N. S. at _London_.

Looking for the 19th of _May_, 1757, in _White_’s Ephemeris, I find that _Jupiter_’s place at that time is in about 12 degrees of ♏; latitude about 1¼ degree North. Then looking for that point upon the Celestial globe, I find that ♃ is then nearly in conjunction with the bright Star in the Southern Balance, and about 1 degree North of it.

To find when he rises and sets, and comes to the meridian: Having put a little black patch on the place of _Jupiter_, elevate the globe according to the latitude, and having brought the Sun’s place to the meridian, set the hour index to 12 at noon; then turn the mark which was made for _Jupiter_, to the Eastern part of the horizon, I find ♃ will rise somewhat more than half an hour after three in the afternoon; and turning the globe about, I find it comes to the meridian a little before eleven at night; and sets almost a quarter past six next morning.

This example being understood, it will be easy to find when either of the other two superior Planets, _viz. Mars_ and _Saturn_, rise, set, and come to the meridian.

I shall conclude this subject about the Globes with the following problems.

PROB. XLVI. _To find all that space upon the
Earth, where an Eclipse of one of the Satellites
of_ Jupiter _will be visible._

Having found that place upon the Earth, in which the Sun is vertical at the time of the eclipse, by _Prob. 13_, elevate the globe according to the latitude of the said place; then bring the place to the meridian, and set the hour index to 12 at noon. If _Jupiter_ be in consequence of the Sun, draw a line with black lead, or the like, along the Eastern side of the horizon, which line, will pass over all those places where the Sun is setting at that time; then count the difference betwixt the right ascension of the Sun, and that of _Jupiter_, and turn the globe Westward, ’till the hour index points to this difference; then keep the globe from turning round its axis, and elevate the meridian, according to the declination of _Jupiter_. The globe being in this position, draw a line along the Eastern side of the horizon; then the space between this line, and the line before drawn, will comprehend all those places of the Earth where _Jupiter_ will be visible, from the setting of the Sun, to the setting of _Jupiter_.

But if _Jupiter_ be in antecedence of the Sun (_i. e._ rises before him) having brought the place where the Sun is vertical, to the zenith, and put the hour index to 12 at noon, draw a line on the Western side of the horizon; then elevate the globe according to the declination of _Jupiter_, and turn it about Eastward, until the index points to so many hours distant from noon, as is the difference of right ascension of the Sun and _Jupiter_. The globe being in this position, draw a line along the Western side of the horizon; then the space contained between this line, and the other last drawn, will comprehend all those places upon the Earth where the Eclipse is visible, between the rising of the Sun, and that of _Jupiter_.

_The_ DESCRIPTION _of the Great_ ORRERY,
_lately made by Mr._ THOMAS WRIGHT,
Mathematical Instrument-Maker to his late
MAJESTY, and now by BENJAMIN
COLE, _his Successor_.

The ORRERY is an Astronomical Machine, made to represent the motions of the Planets. These machines are made of various sizes, some having more Planets than others; but I shall here confine myself to the description of that above-mentioned.

In the Introduction we gave a short account of the _Order_, _Periods_, _Distances_, and _Magnitudes_ of the _Primary Planets_; and of the _Distances_ and _Periodical Resolutions_ of the _Secondary Planets_ round their respective Primaries. We shall here explain their _Stations_, _Regradations_, _Eclipses_, _Phases_, _&c._ but first let us take a general view of the _Orrery_.

[Sidenote: The Description of the _Orrery_.]

[Sidenote: Vide _Frontispiece_.]