Part 8
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.