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4. Gad=the complement of the latitude, being measured by the arc hæ the elevation of the equator above the horizon. Any two of these quantities being given, the rest may
This triangle is complemental to the triangle GÆ zenith ; for Gz is the complement of do the altitude; ÆG is the hour from noon; Æz the latitude; and the angle wzg, measured by the arc hd, is the sun's azimuth from the meridian.
In latitude 510.32' North, when the sun was on the equinoctial, his altitude was 220.15'; required his azimuth, and the hour of the day?
1. To find ad the complement of the azimuth, in the triangle. adg.
Rad x sine Ad=tang dg x cot Gad. :(log tang dg+log cot gad) -10=log sine ad. or, Rad : tang dg = 22°.15'::cot Gad = 389.28' : sine ad = 30°.59'.
Hence the azimuth = 59o.1' from the south towards the east or west, according as the time is before or after noon. 2. To find AG, the hour from 6, in the triangle adg.
Rad x sine dg=sine ag x sine Gad. (10+log sine do)-log sine gad=log sine AG. or, Sine Gad=38°.28' : rad::sine dg = 22°.15': sine ag = 37° 29'.45', which reduced into time=2h.29'.59"; this added to, and subtracted from, 6 hours, gives 8h.29'.59", or 3h.30':1", according as the sun is in the eastern or western semi-circle of the globe.
1. In latitude 510.32' N. when the sun bas no declination, what is his altitude, and azimuth at three hours and a half from noon? Answer. The altitude = 220.15'. Azimuth = 6. 59o.1' E.
. or W.
2. At the time of the equinox the sun's altitude was found to be 22°.15', and his azimuth S. 59° E.; required the hour of the day, and the latitude of the place?
Answer. Time gh.30'. Latitude 510.32 North,
3. In latitude 51°.32' North, the sun being in the equinox, there is given the angle (Agd) which the vertical circle passing
through the sun forms with the equinoctial=570:46'; to find the hour of the day and the altitude and azimuth of the sun ?
Answer. Time 8h.29'.56", or 3h.30'.1". Altitude 22o.15'; azimuth S. 59'. E. or W.
(H) PROBLEM VI. (Plate III. Fig. 2.) . The difference of longitudes between two places, both in one parallel of latitude, being given, to find the distance between them.
Before we give any examples to this problem, it will be necessary to remark, that all the meridians on the terrestrial globe are great circles meeting each other in the poles of the equator. Therefore the meridional distances of places vary as the latitudes; that is, the distance between the meridians PE and pw is greater on the equator woQE (supposing woQE to be a part of the equator) than it is on the parallel of latitude MN, and greater at MN than at li, &c.
Longitude is always counted on the equator, where a degree is reckoned 60 geographical miles, hence a degree of longitude in the parallel mn must be less than 60 miles. This shows that what is called meridional distance in navigation, is always less than the longitude, except the ship sail on the equator.
Since all the meridians PW, PO, PQ, &c. cut the equator at right angles (I. 134.), they must necessarily cut all the parallels of latitude at right angles.
The arcs of the equator wo, 09, &c. are the measures of the spherical angles wPO, OPQ, &c. at the pole (D. 133).
If wm or on represent the latitude of two places on the same parallel, then mp or np will be their co-latitudes.
The arcs wo and mn are similar (P. and R. 135.); hence mr : mn:: cw : wo, viz. cw : mr::wo : mn. But my is the sine of the latitude wm, mr-vc is its cosine, and cw is the radius of the sphere. Therefore,
Radius : cosine of any latitude::the length of any arc on the equator : the length of a similar arc in that latitude.
PRACTICAL EXAMPLES. 1. Suppose a ship, in latitude 40° North, sail due east until her difference of longitude be 60', required the distance sailed in that parallel ? Rad : cos 40°::diff. long=3600 miles : the distance 2757.7 miles.
EXAMPLE II. 2. Suppose a ship, in latitude 40° north, sail due east 2757.7 miles, what is her difference of longitude ?
Anseer. 3600 miles.
(I) By the first of these examples, geographers have calculated tables showing how many miles answer to a degree of longitude for each degree of latitude; thus in latitude 45°, a degree of longitude is 42.43 miles; in the latitude of London 51°.32', the length of a degree is 37.32 miles, &c.
3. If a ship sail 275707 miles directly eastward or westward, so as to alter her longitude 60°, what latitude is she in?
Answer. 40° north or south.
4. Suppose a ship to sail from latitude 40° North to Jatitude 10° North, till her difference of longitude be 60°, and it be required to find her departure, course, and distance sailed.
Here ll the parallel of 40° is the latitude left, aa the parallel of 10' is the latitude arrived at, and mn the parallel of 25° is the middle latitude between them; hence MP is the complement of the middle latitude, and mn is the meridional distance, or the true departure nearly* ; Bw is a quadrant or 90°, and WE is the difference of longitude. (K) Hence the following CONSTRUCTION (Plate III. Fig. 3.):
1. With the chord of 60', which is equal to the sine of 90°, and centre P, describe the arc we, on which set off the difference of longitude (60°= 3600 miles); draw pw and PE.
2. With the sine of (65°) the complement of the middle latitude, taken from the line of sines on Gunter's scale, describe the arc Mn, and draw the straight line MN.
3. Set off the difference of latitude (30°=1800 miles) from P to m; draw mn perpendicular to pm, and equal to MN; lastly, draw pn, and the construction is finished. The several lines are named as in the figure.
(L) OR THUS. Make the angle MPN (Plate III. fig. 4.)=the complement of the middle latitude, and on the difference of longitude, then MN will be the departure or meridional distance ; for, in the plane triangle MPN, per Plane Trigonometry, radius : PN:: sine 2 MPN:
MN, the same proportion as is used in the first example. The triangle mpn is the same as before.
(M) OR THUS. (Plate III. Fig. 5.) Draw the meridian Ap, in which assume any point P; make the angle at P equal to the complement of the middle latitude, in the opposite quadrant to which the ship sails; set off pnequal to the difference of longitude, and draw mn perpendicular to
AP; lastly, set off the difference of latitude from a to d, and draw An; then the names of the several lines are as expressed in the figure.
BY CALCULATION. (Plate III. Fig. 3, 4, or 5.) Rad : diff. long::sine comp. midd. lat. : merid. dist. or departure=3262:7.
Diff. lat : rad:: departure : tang course=61°.7.
THE APPLICATION OF OBLIQUE ANGLED SPHERICAL
TRIANGLES TO ASTRONOMICAL PROBLEMS.
PROBLEM VII. (Plate III. Fig. 1.) (N) Given the sun's declination, and the latitude of the place, to find the apparent time of day-break in the morning, and the end of twilight in the evening.
1. In the oblique-angled spherical triangles Ozenith n. 1. On=the sun's distance from the north pole.
2. Ozenith the sun's distance from the zenith. . 3. Zenith n=the complement of the latitude.
4. On zenith, measured by the arc ax=the time from noon. OR, 2. In the oblique-angled spherical triangle onadir s.
1. O s=the sun's distance from the south pole. 2. O Nadir the sun's distance from the nadir. 3. s Nadir=the complement of the latitude.
4. Os Nadir, measured by the arc ag,=the time from midnight.
EXAMPLE. Given the latitude of the place 51°.32 N., and the sun's declination 10° N., to find the time of day-break in the morning, and the end of twilight in the evening.
In the triangle o zenith n. ON=80° the sun's distance from the north pole=90° — 10°.
O zenith = 108° the sun's distance from the zenith = 18° +90°.
Zenith n=38o.28' the complement of the latitude.
To find the angle zenith NO, measured by the arc ax, = the time from noon. ON
Cosec zn = 389.28'- '206177 reject Ozenith = 108° Cosecon=80°. 0'- .00665 ) indices. Zenith N= 38°.28' || Sine 113o.14' 9.96327
5o.14' 8.96005 2 | 226. 28
2 | 19:13614 Half sum
113. 14 Zenith=108. O Cosine 68o.17.29" 9.56807
2 Remainder 5. 14
LONZ=1369.34'.58"=9h.6.20", time from noon* when the sun is 18° below the horizon. Conse, quently the day breaks at 25.53'.40" in the morning, and twilight ends at gh.6'.20" in the evening, supposing the sun's declination to undergo no change between the beginning of twilight in the morning, and the ending thereof at night, being about 18 hours.
The same things might have been found from the triangle Os nadir, for so=90° +10°=100°, Nadir = 180°-108° = 72°, and Nadir s = comp. lat. = 38o.28. Then by the method above find the angle O s nadir (measured by the arc ag)=43o.25'.12"=25.53.40" as before, the time from midnight, when the sun is 18° below the horizon.
(O) When the declination of the sun, the latitude and declination being of the same name, is greater than the difference between the complement of latitude and 18°, the parallel of declination ( sss) will not cut the parallel of 18o (TO) below the horizon: consequently there will be no real night at these times, but constant day or twilight, as is the case at London from the 22d of May to the 21st of July.
(P) Since the sun sets more obliquely at some times of the year than at others, it necessarily follows that he will be longer in descending 180 below the horizon at one season than at another.
When the sun is on the same side of the equinoctial as the visible pole, the duration of twilight will constantly increase as he approaches that pole, till be enters the tropic, at which time
* The O nz represents the time from apparent noon, that is, from the time the comes to the meridian, and the time from 12 o'clock, as shown by a well regulated clock or time-piece. If the equation of time, given in the second page of each month in the Nautical Almanac, be applied to this apparent time by addition or subtraction, as there directed, you will obtain the true time, or that shown