the elevated pole, PN" is about 66°, and, consequently, PN” — 18° — 48°. Hence, at a place whose latitude is more than 48°, the twilight continues all night at the time the sun's declination is greatest and of the same name with the latitude. PROBLEMS. 183, To find the latitude of a place. = 1st Method. Let M, M', m or m', Fig. 31, be the point in which the sun or a fixed star passes the meridian PRN. Then we have the latitude ZQ = ZMQM ZM' QM = Qm-Zm=Qm' - Zm'. Hence, calling the zenith distance of the body north or south according as the zenith is north or south of the body, when the declination of the body and its correct meridian zenith distance are of the same name, their sum will be the latitude, which will be of that name; and when they are of different names, their difference will be the latitude of the same name with the greater quantity; observing however that when the body passes the meridian below the pole, the supplement of the declination must be used instead of the declination itself. Consequently, when from the observed meridian altitude of the sun or a star, the correct altitude has been found, by applying the proper corrections, the latitude is thus very easily obtained. If two stars be selected, one of which passes the meridian to the south of the zenith and the other to the north, at about the same latitude, and the latitude be obtained by each, the mean of the two results will be nearly free from any small errors depending on want of accuracy in the centering or adjustment of the instrument used in observing the altitudes, or in the table of refractions. For as such errors would affect the observed altitudes equally or nearly so, making them both too great or too small, by the same quantity, it is obvious, from the expressions for the latitude in the two cases, that the latitude obtained by one star must be as much too great, as that obtained by the other is too small. 2nd Method. Let S, Fig. 33, be the position of a star out of the meridian, and let SD be an arc of a great circle perpendicular to the meridian. If the altitude of the star be observed and corrected for refraction, and the time at which the altitude is taken, be also observed, we shall have given in the triangle ZPS, the two sides PS and ZS and the angle ZPS, to find PZ, the complement of the latitude. For PS is known from the declination of the star, ZS is the complement of the correct altitude, and the angle ZPS, the star's distance from the meridian, is the difference between the star's right ascension and the sidereal time of observation, expressed in degrees. If the observed time is mean solar time, the corresponding sidereal time must be obtained.* (142). In the right angled triangle PDS we have (App. 49), tang PD = cos SPD tang PS. And, from the right angled triangles PDS and ZDS, we have (App. 45), or, cos PS: cos PD :: cos ZS : cos ZD. The difference between PD and ZD, or their sum, when D falls between P and Z, gives PZ, the complement of the latitude. It is best to make the observations when the star is near the meridian, as a slight inacccuracy in the observed time, does not then sensibly affect the computed latitude. This is not, however, material when it is the Pole star that is observed, as its motion in altitude is at all times, slow. The star selected should not be one that passes the meridian so near the zenith as to leave a doubt with regard to the side of it, on which the perpendicular SD would fall. There are various other methods of finding the latitude of a place; one of which has been given in a previous article (58). In the Nautical Almanac, the sidereal time at mean noon, at Greenwich, is given for cach day in the year; and the method of finding it, for any time, at any meridian, is given also. 183. Given the latitude of a place and the sun's declination, to find the time of his rising or setting. Let HWR, Fig. 31, be the western half of the horizon, Z the zenith, EQ the equator, and P the elevated pole. Also let NM be parallel to EQ, at a distance equal to the given declination. Then will A, its intersection with HR, be the point of the horizon at which the sun sets, and the hour angle APM, converted into time, will be the interval between noon and sunset. Supposing the declination not to change during the day, it is evident the interval between sunrise and noon will be just the same. The angle APM or its measure, the arc DQ, is called the semi-diurnal arc, and the arc ED, the semi-nocturnal arc. The arc DW, contained between a declination circle, passing through the point of setting or rising, and the western or eastern point of the horizon, is called the ascensional difference. In the right angled spherical triangle ADW, we have (App. 48), = tang AD = sin DW tang EWH - sin DW cot PH, tang PH tang AD, = tang lat. x tang declin. Now, as the angle WPQ or its measure WQ, is 90° or 6 hours, it is evident that when the declination is of the same name with the latitude, the sun will set at some point A, between H and W, and the ascensional difference DW, converted into time and added to 6 hours, will give the semidiurnal arc in time. When the declination is of a different name from the latitude, the sun will set at some point A', between W and R, and the ascensional difference subtracted from 6 hours, will give the semi-diurnal arc. The semi-diurnal arc in time, evidently, expresses the time of sunset, and subtracted from 12 hours, it gives the time of sunrise. These are in apparent time, which may be converted into mean, by applying the equation of time. We may find an expression for the semi-diurnal arc without first obtaining the ascensional difference. For cos DQ=cos (WQ + DW) = cos (90° + DW)=sin DW. 184. To find the time of the sun's apparent rising or setting. When the sun's centre appears to be in the horizon at a point a, Fig. 31, we have (80 and 91), his zenith distance Zb = 90° + refraction-parallax. Put, L=HP = latitude of place, DPb 90° sun's declination, Rrefraction P = angle aPZdQ semi-diurnal arc. Then in the spherical triangle Z¿P, we have (App. 38), sin ZPb = √ ± or, sin | P= = sin (Zb+Pb — PZ) sin † ( Zb + PZ— Pb sin Pb sin PZ ‚sin 1⁄2 (90° +R+D—90°+L)sin † (90° +R+90°—L—D) sin D sin (90°-L) If we take R = refraction-parallax-semi-diameter, the above formulæ gives the semi-diurnal arc for the apparent rising or setting of the sun's upper limb. 185. To find the time of beginning or end of twilight. At the beginning or end of twilight the sun is 18° below the horizon (88). Let B be the position of the sun when at this distance below the horizon. Then, in the triangle ZBP, we have PZ = 90° L, PB = PA = D, ZB = Hence, as in the last article, 90° 18°, and BPM P. 14 sin (D+L+ 18°) cos (D+ L-18°) sin D cos L 186. Given the latitude of a place and the sun's declination and altitude, to find the time of day. Let S be the position of the sun. Taking D and L as above, put H = KS = sun's altitude, and P = hour angle SPZ. Then, ZS ZK-KS — 90° — H. 90° H. in article 184, Hence, as sin ZPS = ✔ sin † (ZS + PZ — PS) sin † (ZS + PS — PZ), sin PS sin PZ sin 1⁄2 (90°—H+90°—L—D)sin } (90°—H+D—90°+L), sin D sin (90°- L) or, sin P=✔ 187. Conjunction &c. A body is said to be in conjunction with the sun, or simply, in conjunction, when its position is such, that its longitude and that of the sun are the same;* to be in opposition, when their longitudes differ 180°; and to be in quadrature, when their longitudes differ 90° or 270°. The term syzygy is used to denote either conjunction or opposition. When the body is in any of the four positions, midway intermediate between the syzygies and quadratures, it is said to be in octant. As each of the planets Mercury and Venus, revolve at a less distance from the sun than that of the earth (18), either of them may be in conjunction both on the same side of the sun with the earth and on the opposite side. The former is called inferior, and the latter, superior, conjunction. Some of the preceding terms are frequently designated by characters as follows: Conjunction 6 Opposition 8 Quadrature . * When any two of the heavenly bodies have the same longitude, they are said to be in conjunction. |