EXAMPLES. 1. On the 1st of August, 1831, the sun's declination was 18° 10′ 22′′ N., at what hour was he due east at Greenwich, and what was his altitude at that time? Hence the time is 35 seconds and a half past 7 o'clock, and the altitude 23° 29′ 37′′. 2. Given the sun's declination 5° 8′ 26′′ N., and his altitude when due east 16° 53′ 10′′; required the latitude of the place. Latitude 17° 58′ N. 3. If the declination of a celestial object be 18° 4' S., what is its altitude when on the prime vertical of latitude 27° 42′ S., and its distance from the meridian in time? Altitude 41° 51'; Merid. distance in time 3h 26m 20s. PROBLEM V. To find the time when the apparent motion of a celestial object is perpendicular to the horizon, from having its declination and the latitude of the place given. Let s s' represent the parallel of declination, or the apparent diurnal path of the body, and let the vertical ZSN be drawn to touch it in S; then S will be the place of the body when its apparent motion is in the direction SZ of the vertical, and therefore perpendicular to the horizon: through S draw the hour circle PS, which, being the shortest distance from P to ZN, is perpendicular to it (p. 85); hence the triangle PSZ is right-angled at S, and in which we have given the colatitude ZP, and the codeclination PS, to find the hour angle ZPS. It is obvious that this problem will be impossible when Ps' exceeds PZ; that is, when the declination is less than the latitude. PROBLEM VI. To determine upon what vertical a celestial object must be, in order that a small error, committed in taking its altitude, may have the least possible effect upon the hour angle. S Let S be the place of the sun, or other body, but by an error in taking its altitude let it be referred to S'. Draw S′ S" parallel to the horizon, and meeting the parallel of declination s s' in S", then when the body is at S" it will really have the coaltitude ZS": = ZS', which it was erroneously supposed to have at S, so that in the determination of the hour angle P, from the colatitude, the coaltitude, and the codeclination, the small angle S' PS" will be the amount of the error. As the triangle SS' S" is, of course, exceedingly small, it may be regarded as a rectilinear triangle, right-angled at S'; therefore SS' = SS" sin. S", and SS" =sin. PS / SPS", (see p. 121,) consequently, SS' SS' sin. S" sin. PS SPS" ... SPS" = sin. S" sin. PS (1); Now the angle S" is equal to the angle ZSP, because S' SS" is the complement of each, and therefore, by the relation between the sides and angles of a spherical triangle, we have Substituting the second member of this equation in (1), we have SS' error in alt. SPS' = sin. PZ sin. SZP cos. lat. sin. azimuth This expression will obviously be the least possible when the sine of the azimuth is the greatest possible, or when the azimuth is 90°; that is, when the body is on the prime vertical. Hence, in deducing the time from an altitude of any celestial body, it will be best to make the observation when the body is either exactly, or nearly, due east or due west. PROBLEM VII. The latitudes and longitudes of two celestial objects being given, to determine their distance apart. Let P represent the pole of the ecliptic, and PS, PS', two arcs of celestial latitude, drawn to the two objects S, S'; then will these arcs represent the colatitudes, the angle P will be the difference of longitude, and the arc SS' will be the distance sought, so that we have two sides and their included angle given to find the third side. In order to this we must first determine, agreeably to the method explained at page 99, a subsidiary angle, by the equation cot. tan. PS cos. P; after which the side SS' is found by the equation EXAMPLES. 1. Required the distance between Procyon and Capella, the latitude of Procyon being 15° 58′ 14′′ S., and its longitude 3o 22° 55′ 42′′; also the latitude of Capella being 22° 51′ 57′′ N., and longitude 2o 1 8° 57' 57"? Taking the difference of the longitudes, we have for the angle P, P=33° 57′ 45′′; and for the colatitudes we have PS = 105° 58′ 14′′, PS′ =67° 8′ 3′′; hence the logarithmic process will be as follows: In this example cot. w is negative, because tan. PS is negative, and cos. P positive; also cos. SS' is positive, because cos. PS is negative, sin. (w+PS') negative, and sin. w positive. The operation will obviously be similar, when, instead of the latitudes and longitudes, the right ascensions and declinations of the two bodies are given to find their distance apart. 2. The latitude and longitude of a star S. are 38° 40′ 26′′ N., and 3$ 2° 4′ 40′′; and of a star S'., 13° 26′ 11′′ N., and 9° 11° 41′ 26′′; required their angular distance apart. Distance 127° 7′ 11′′. 3. What is the distance between Sirius and Procyon, the right ascension of Sirius being 99° 0′ 21′′, and its declination 16° 26′ 35′′ S.; and the right ascension of Procyon 112° 6' 47", and its declination 5° 45' 3" N. PROBLEM VIII. Given the latitude of the place and the sun's declination to find the beginning and end of twilight. Twilight commences in the morning and ends in the evening, when the sun is about 18° below the horizon. Hence, if PZ (see the diagram to next problem,) represent that portion of the meridian which is intercepted between the elevated pole and the zenith, and S′ be that point in the sun's apparent path on any day which is 108° from Z, S′ will be the place of the sun at the commencement of morning twilight, or at the termination of evening twilight; also PS' will be the codeclination, and PZ the colatitude; we thus have the three sides of the triangle PS'Z, to find the angle P. Hence, calling the sum of the three sides S, the formula for computing the hour angle P will be sin. P= sin. (SZP) sin. ( S — PS') sin. ZP sin. PS' ; which is the same as sin. P= sin. (lat. +18° + codec.) sin. (dec. +18° +colat.), cos. lat. cos. dec. a very convenient form for computation. ; EXAMPLES. 1. At what time did twilight commence at Edinburgh, lat. 55° 57' 20" N., on the 20th of August, 1831, when the sun's declination was 12° 38′ 9′′ N.? |