H. M.
{ _Rome_ 0 52 P. M.
At { _Constantinople_ 2 07 P. M.
{ _Vera-Cruz_ 5 30 A. M.
{ _Pekin_ in _China_ 7 50 P. M.
PROB. XI. _The Day of the Month being given,
to find 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, bring the same to the
meridian, and note the degree over it; then turning the globe round,
all places that pass under that degree will have the Sun vertical that
day.
PROB. XII. _A place being given in the_ Torrid
Zone, _to find those two Days in which the Sun
shall be Vertical to the same._
Bring the given place to the meridian, and mark what degree of latitude
is exactly over it; then turning the globe about its axis, those two
points of the ecliptic, which pass exactly under the said mark, are the
Sun’s place; against which, upon the wooden horizon, you’ll have the
days required.
PROB. XIII. _To find where the Sun is Vertical at
any given time assigned; or the Day of the Month
and the Hour at any Place_ (_suppose_ London) _being
given, to find in what place the Sun is Vertical at
that very time._
Having found the Sun’s declination, and brought the first place
(_London_) to the meridian, set the index to the given hour, then turn
the globe about until the index points to XII at noon; which being
done, that place upon the globe which stands under the point of the
Sun’s declination upon the meridian, has the Sun that moment in the
Zenith.
PROB. XIV. _The Day, and the Hour of the
Day at one place, being given; to find all those
places upon the Earth, where the Sun is then
Rising, Setting, Culminating (or on the meridian)
also where it is Day-light, Twilight, Dark Night,
Midnight; where the Twilight then begins, and where
it ends; the height of the Sun in any part of the
illuminated hemisphere; also his depression in the
obscure hemisphere._
Having found the place where the Sun is vertical at the given hour,
rectify the globe for that latitude, and bring the said place to the
meridian.
Then all those places that are in the Western semicircle of the
horizon, have the Sun rising at that time.
Those in the Eastern semicircle have it setting.
To those who live under the upper semicircle of the meridian, it is 12
o’clock at noon. And,
Those who live under the lower semicircle of the meridian, have it at
midnight.
All those places that are above the horizon, have the Sun above them,
just so much as the places themselves are distant from the horizon;
which height may be known by fixing the quadrant of altitude in the
zenith, and laying it over any particular place.
In all those places that are 18 degrees below the Western side of the
horizon, the twilight is just beginning in the morning, or the day
breaks. And in all those places that are 18 degrees below the Eastern
side of the horizon, the twilight is ending, and the total darkness
beginning.
The twilight is in all those places whose depression below the horizon
does not exceed 18 degrees. And,
All those places that are lower than 18 degrees, have dark night.
The depression of any place below the horizon is equal to the altitude
of its _Antipodes_, which may be easily found by the quadrant of
altitude.
PROB. XV. _The Day of the Month being given;
to show, at one view, the length of Days and Nights
in all places upon the Earth at that time; and to
explain how the vicissitudes of Day and Night are
really made by the motion of the Earth round her
axis in 24 hours, the Sun standing still._
The Sun always illuminates one half of the globe, or that hemisphere
which is next towards him, while the other remains in darkness: And if
(as by the last problem) we elevate the globe according to the Sun’s
place in the ecliptic, it is evident, that the Sun (he being at an
immense distance from the Earth) illuminates all that hemisphere, which
is above the horizon; the wooden horizon itself, will be the circle
terminating light and darkness; and all those places that are below it,
are wholly deprived of the solar light.
The globe standing in this position, those arches of the parallels
of latitude which stand above the horizon, are the _Diurnal Arches_,
or the length of the day in all those latitudes at that time of the
year; and the remaining parts of those parallels, which are below the
horizon, are the _Nocturnal Arches_, or the length of the night in
those places. The length of the diurnal arches may be found by counting
how many hours are contained between the two meridians, cutting any
parallel of latitude, in the Eastern and Western parts of the horizon.
In all those places that are in the Western semicircle of the horizon,
the Sun appears rising: For the Sun, standing still in the vertex (or
above the brass meridian) appears Easterly, and 90 degrees distant from
all those places that are in the Western semicircle of the horizon;
and therefore in those places he is then rising. Now, if we pitch upon
any particular place upon the globe, and bring it to the meridian,
and then bring the hour index to the lower 12, which in this case,
we’ll suppose to be 12 at noon; (because otherwise the numbers upon
the hour circle will not answer our purpose) and afterwards turn the
globe about, until the aforesaid place be brought to the Western side
of the horizon; the index will then shew the time of the Sun rising
in that place. Then turn the globe gradually about from West to East,
and minding the hour index, we shall see the progress made in the day
every hour, in all latitudes upon the globe, by the real motion of the
Earth round its axis; until, by their continual approach to the brass
meridian (over which the Sun stands still all the while) they at last
have noon day, and the Sun appears at the highest; and then by degrees,
as they move Easterly the Sun seems to decline Westward, until, as the
places successively arrive in the Eastern part of the horizon, the Sun
appears to set in the Western: For the places that are in the horizon,
are 90 degrees distant from the Sun. We may observe, that all places
upon the Earth, that differ in latitude, have their days of different
length (except when the Sun is in the equinoctial) being longer or
shorter, in proportion to what part of the parallels stands above the
horizon. Those that are in the same latitude, have their days of the
same length; but have them commence sooner or later, according as the
places differ in longitude.
PROB. XVI. _To explain in general the
alteration of Seasons, or length of the Days
and Nights made in all places of the World, by
the Sun’s (or the Earth’s) annual motion in the
Ecliptic._
It has been shewed in the last problem, how to place the globe
in such a position as to exhibit the length of the diurnal and
nocturnal arches in all places of the Earth, at a particular time:
If the globe be continually rectified, according as the Sun alters
his declination, (which may be known by bringing each degree of the
ecliptic successively to the meridian) you’ll see the gradual increase
or decrease made in the days, in all places of the World, according as
a greater or lesser portion of the parallels of latitude, stands above
the horizon. We shall illustrate this problem by examples taken at
different times of the year.
1. Let the Sun be in the first point of ♋ (which happens on the 21st of
_June_) that point being brought to the meridian, will shew the Sun’s
declination to be 23½ degrees North; then the globe must be rectified
to the latitude of 23½ degrees; and for the better illustration of the
problem, let the first meridian upon the globe be brought under the
brass meridian. The globe being in this position, you’ll see at one
view the length of the days in all latitudes, by counting the number
of hours contained between the two extreme meridians, cutting any
particular parallel you pitch upon, in the Eastern and Western part
of the horizon. And you may observe that the lower part of the arctic
circle just touches the horizon, and consequently all the people who
live in that latitude have the Sun above their horizon for the space
of 24 hours, without setting; only when he is in the lower part of
the meridian (which they would call 12 at night) he just touches the
horizon.
To all those who live between the arctic circle and the Pole, the Sun
does not set, and its height above the horizon, when he is in the
lower part of the meridian, is equal to their distance from the arctic
circle: For example, Those who live in the 83d parallel have the Sun
when he is lowest at this time 13½ degrees high.
If we cast our eye Southward, towards the equator, we shall find, that
the diurnal arches, or the length of days in the several latitudes,
gradually lessen: The diurnal arch of the parallel of _London_ at this
time is 16½ hours; that of the _Equator_ (is always) 12 hours; and so
continually less, ’till we come to the _Antarctic Circle_, the upper
part of which just touches the horizon; just those who live in this
latitude have just one sight of the Sun, peeping as it were in the
horizon: And all that space between the antarctic circle and the South
Pole, lies in total darkness.
If from this position we gradually move the meridian of the globe
according to the progressive alterations made in the Sun’s declination,
by his motion in the ecliptic, we shall find the diurnal arches of
all those parallels, that are on the Northern side of the equator,
continually decrease; and those on the Southern side continually
increase, in the same manner as the days in those places shorten and
lengthen. Let us again observe the globe when the Sun has got within
10 degrees of the equinoctial; now the lower part of the 80th parallel
of North latitude just touches the horizon, and all the space betwixt
this and the pole, falls in the illuminated hemisphere: but all those
parallels that lie betwixt this and the arctic circle, which before
were wholly above the horizon, do now intersect it, and the Sun
appears to them to rise and set. From hence to the equator, we shall
find that the days have gradually shortened; and from the equator
Southward, they have gradually lengthened, until we come to the 80th
parallel of the South latitude; the upper part of which just touches
the horizon; and all places betwixt this and the South Pole are in
total darkness; but those parallels betwixt this and the antarctic
circle, which before were wholly upon the horizon, are now partly
above it; the length of their days being exactly equal to that of the
nights in the same latitude in the contrary hemisphere. This also holds
universally, that the length of one day in one latitude North, is
exactly equal to the length of the night in the same latitude South;
and _vice versa_.
Let us again follow the motion of the Sun, until he has got into the
equinoctial, and take a view of the globe while it is in this position.
Now all the parallels of latitude are cut into two equal parts by the
horizon, and consequently the days and nights are of equal lengths,
_viz._ 12 hours each, in all places of the world; the Sun rising and
setting at six o’clock, excepting under the two _Poles_, which now lie
exactly in the horizon: Here the Sun seems to stand still in the same
point of the heavens for some time, until by degrees, by his motion in
the ecliptic, he ascends higher to one and disappears to the other,
there being properly no days and nights under the Poles; for there the
motion of the Earth round its axis cannot be observed.
If we follow the motion of the Sun towards the Southern tropic, we
shall see the diurnal arches of the Northern parallels continually
decrease, and the Southern ones increase in the same proportion,
according to their respective latitudes; the North Pole continually
descending, and the South Pole ascending, above the horizon, until the
Sun arrives into ♑, at which time all the space within the antarctic
circle is above the horizon; while the space between the arctic circle,
and its neighbouring Pole, is in total darkness. And we shall now find
all other circumstances quite reverse to what they were when the Sun
was in ♋; the nights now all over the world being of the same length
that the days were of before.
We have now got to the extremity of the Sun’s declination; and if we
follow him through the other half of the ecliptic, and rectify the
globe accordingly, we shall find the seasons return in their order,
until at length we bring the globe into its first position.
The two foregoing problems were not, as I know of, published in any
book on this subject before; and I have dwelt the longer upon them,
because they very well illustrate how the vicissitudes of days and
nights are made all over the world, by the motion of the Earth round
her axis; the horizon of the globe being made the circle, separating
light and darkness, and so the Sun to stand still in the vertex. And
if we really could move the meridian, according to the change of the
Sun’s declination, we should see at one view, the continual change made
in the length of days and nights, in all places on the Earth; but as
globes are fitted up, this cannot be done; neither are they adapted
for the common purposes, in places near the equator, or any where in
the Southern hemisphere. But this inconvenience is now remedied (at a
small additional expence) by the hour circle being made to shift to
either Pole; and some globes are now made with an hour circle fixed to
the globe at each Pole between the globe and meridian, so as to have
none without side to interrupt the meridian from moving quite round the
wooden horizon.
PROB. XVII. _To shew by the globe, at one
view, the longest of the Days and Nights in any
particular places, at all times of the Year._
Because the Sun, by his motion in the ecliptic, alters his declination
a small matter every day; if we suppose all the torrid zone to be
filled up with a spiral line, having so many turnings; or a screw
having so many threads, as the Sun is days in going from one tropic
to the other: And these threads at the same distance from one another
in all places, as the Sun alters his declination in one day in all
those places respectively: This spiral line or screw will represent
the apparent paths described by the Sun round the Earth every day; and
by following the thread from one tropic to the other, and back again,
we shall have the path the Sun seems to describe round the Earth in a
year. But because the inclinations of these threads to one another are
but small, we may suppose each diurnal path to be one of the parallels
of latitude, drawn, or supposed to be drawn upon the globe. Thus much
being premised, we shall explain this _Problem_, by placing the globe
according to some of the most remarkable positions of it, as before we
did for the most remarkable seasons of the year.
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.