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,Maythe 21st atLondonwhen 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.
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.
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, andits declination 11½ degrees North, the latitude will be 51½ degrees North.
[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, BecauseLunareclipses 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 ofthe 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.
1. To find theBabylonic 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 theBabylonic Hour.
2. To find theItalic 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 theItalic Hourrequired.
3. To find theJewish Hour(which is ¹/₁₂ part of anArtificial 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 theJudaicalhour required.
Thus, if the Sun rises at 4 o’clock (consequently sets at 8) and the hour given be 5 in the evening, theBabylonishhour will be the 13th, theItalicthe 21st and theJewishhour will be nine and three quarters.
The converse being given, the hour of the day, according to our way of reckoning inEngland, may be easily found.
The following Problems are peculiar to theCelestial Globe.
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 theSun’s Right Ascension; and that degree of the meridian, which is exactly over the Sun’s place, is theSun’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 theTerrestrial 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].
Note, The declination of the Sun may be found after the same manner by theTerrestrial 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 inAstronomy, 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.
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 ofArcturuswill be found to be 31 degrees North,and the longitude 200 degrees from ♈, or 20 degrees from ♎: The latitude ofFomalhautin 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 theTerrestrialGlobe, inProb.VII.
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 whosebearing 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 thePole 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 ofquadrangle; these seven Stars make the constellation called theLittle Bear; the Pole-Star being in the tip of the tail. In this neighbourhood you will observe seven bright Stars, which are commonly calledCharles’s Wane; these are the bright Stars in theGreat Bear, and form much such another figure with those before-mentioned in thelittle Bear: The two foremost of the square lie almost in a right line with the Pole Star,and are called thePointers, 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 inquadrangles,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.
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 theOblique Ascension; and where the given point cuts the horizon, is theAmplitude Ortive: If the globe be turned about until the given point be brought to the Western side of the horizon, it will there show theAmplitude Occasive; and where the horizon cuts the equinoctial at that time, is theOblique 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 theAscensional 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 theSemi-diurnal Arch;the complement whereof to a semicircle, is theSemi-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 inProb. 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, atLondon, on the 31st ofMay, theSun’s
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.
Having rectified the globe for the latitude of the Sun’s place, bringthe given Star to the meridian, and also to the East or West side of the horizon, and the index will shew accordingly when the Starculminates, or the time of therisingorsetting.
Thus atLondon, on the 21st ofJanuary,Syriuswill 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 whenSyriuscomes to the meridian at 10½ hours after noon, the Sun’s place will be ≈ ¼ deg.
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.
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 intersectiondescribe 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.
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.
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 theLatitudesought.
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 theequinoctial colureto 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 theHourof the day, and the quadrant cut theAzimuthin 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.
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 Starrises 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 Starsets Cosmically. So likewise against the degree which sets with the Star, you will find theday of the month of theAchronical 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 Starrises Achronically.
Thus, in the latitude ofLondon,Syrius, or theDog-Star, risesCosmicallythe 10th ofAugust, and setsCosmicallythe 10th ofOctober.Aldebaran, or theBull’s Eye, risesAchronicallyon the 22d ofMay, and setsAchronicallyon the 19th ofDecember.
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 thefirst magnitude, when sherises 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 Starrises Heliacally. To find theHeliacal 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 Starsets Heliacally.
Thus you will find thatArcturusrises Heliacally the 28th ofSeptember, and sets HeliacallyDecemberthe 2d.
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 markthat 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 ofJupiteramong the Fixed Stars in the heavens, and also what time he rises and sets, and comes to the meridian on the 19th ofMay, 1757, N. S. atLondon.
Looking for the 19th ofMay, 1757, inWhite’s Ephemeris, I find thatJupiter’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 ofJupiter, 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 forJupiter, 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. MarsandSaturn, rise, set, and come to the meridian.
I shall conclude this subject about the Globes with the following problems.
Having found that place upon the Earth, in which the Sun is vertical at the time of the eclipse, byProb. 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. IfJupiterbe 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 ofJupiter, and turn the globeWestward, ’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 ofJupiter. 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 whereJupiterwill be visible, from the setting of the Sun, to the setting ofJupiter.
But ifJupiterbe 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 ofJupiter, 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 andJupiter. 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 ofJupiter.