12° 31'.23" Hence 90° -12°.31.23"=770.28.37"Od the azimuth from the north. EXAMPLE II. At what time will Arcturus appear upon the six o'clock hour line at Greenwich, latitude 510.28'.40" N., and what will its altitude and azimuth be on the first of April 1822? Its right ascension being 14h.7.32" and declination 20°.6'.50" N. (Table VIII.), and the sun's right ascension on the first of April (Naut. Alm.) being 06.41'.11", and on the second of April 06.44'.49". 1. To find the time of the star's culminating. By Problem VII. (D. 273.) Arcturus will culminate at 13h.24'.19". Consequently (B. 289.) it will be upon the six o'clock hour circle at (13b.24.19" —65=) 7h.24'.19" in the evening, in the eastern hemisphere; and at (13h.24'.19"+6h--) 19h 24'.19", or 76.24'.19" the next morning, in the western hemisphere. 2. To find the altitude ds, in the triangle ads. Rad x sine ds=sine SAD x sine As. (log sine sad #log sine as) –10=log sine ds. or, Rad: sineas=20°.6'.50": :sine sad=519.28'.40": sine Ds= 159.36'.25". 3. To find the azimuth ad from the east, in the triangle ads. Rad X COS SAD=cot as x tang ad. :: (10+ log cos saD) – log cot as= log tang ad. or, Cotas=20°.6'.50" : rad::cos SAD=519.28'.40" : tang AD= 120.50'.56". Hence od=770.9'.4" the azimuth from the north. (C) On account of the small change in the right ascension and declination of a star, it may, without material error, be said to have the same altitude and azimuth every time it arrives at the six o'clock hour circle; and the difference of the times it arrives there, may be considered as equal to the diurnal difference of the sun's right ascension, viz. about four minutes. (Z. 288.) PRACTICAL EXAMPLES. 1. At what time will Aldebaran appear upon the six o'clock hour circle at Greenwich, latitude 519.25.40' N., and what will its altitude and azimuth be on the 20th November 1822 ? 11 The right ascension being 45.25'43";and declination 16°.8'.36" N. (Table VIII.); also, the sun's right ascension on the 20th November (Naut. Alm.)= 151.41.29" andon the 21st 154.45'.40", Answer. Aldebaran culminates at 12h.42 (D. 273.), therefore it is on the 6 o'clock hour circle in the eastern hemisphere at 6h.42' in the evening; and in the western hemisphere at 6h.42 the next morning. Its altitude at each time being 120.32'.3", and the azimuth 799.46.50% from the north. 2. Required the sun's altitude and azimuth at 6 o'clock, in latitude 510.32' N. when his declination is 150.28' North? Answer. Altitude=119.59'.17", azimuth 800.17.14" from the North. 3. Given the sun's declination 199.39' North, his altitude at 6 equal 15o.16'; required the azimuth and the latitude? Answer. Azimuth=770.28'.37" from the elevated pole, latitude=519.32 North. 4. Given the sun's declination=190.39' North, his azimuth at 6 in the morning N. 77o.28.37" E.; required his altitude and the latitude ? Answer. Altitude=150.16', latitude= 51°32' North, 5. At what time will Regulus appear upon the six o'clock hour circle NsaS at Greenwich, latitude 519.28'.40' North, and what will its altitude and azimuth be on the 6th February 1822? The right ascension being gh.58'.53" and declination 12°, 50'.1" N. (Table VIII.), also the sun's right ascension at noon on the same day (Naut. Alm.)=21h.18'41", and on the 7th February 216.22.41"? Answer. Regulus culminates at 125.38' (D. 273.), is on the 6 o'clock hour circle, at 6938' in the evening in the eastern hemisphere; and at 6h.38' the next morning in the western hemisphere. Its altitude=100.1'.11", and its azimuth=81o. 55'.28" from the north. 6. At what time will Castor appear upon the 6 o'clock hour circle at Greenwich, latitude 51°.28'.40' N., and what will its altitude and azimuth be on the 1st of December 1822 ? The right ascension being 7h.23'.13" and declination 32o.16' 10" N. (Table VIII.), also the sun's right ascension on the first of December at Noon (Naut. Alm.) = 166.28'.16", and on the second 165.32.35". Answer. Castor culminates at 146.52'.16", (D. 273.) will be on the six o'clock hour circle at 8h.52.16" in the evening, in the eastern hemisphere, and at 8h.52.16" next morning in the western hemisphere; altitude=249.41'.21", azimuth=680.31'. 55" from the north. (D) PROBLEM IV. (Plate III. Fig. 1.) The latitude of a place, and the declination of the sun (or of a star) being given; to find the altitude, and the time when it revill be due east and west. In the right-angled spherical triangle ags. 1. Ag=the time from six o'clock when the sun is due east or west. Or it is the complement of the time from a star's culminating 2. As the altitude of the sun, or a star, above the horizon, when on the prime vertical zsan. -3. Gs=the sun's or star's declination, of the same name as the latitude. 4. The angle gas=the latitude of the place; for the elevation Hw of the equinoctial is always equal to the complement of the latitude. Any two of these quantities being given, the rest may be found. The triangle sn Zenith, right-angled at Zenith, is complemental to the triangle ags. EXAMPLE I. In latitude 510.32' North, when the sun's declination is 199.39' North, required his altitude, and the time when he will appear due east or west? 1. To find the altitude as in the triangle AGs. Rad x sine .gs=sine Gas x sine As. :: (10 +log sine Gs) - log sine gas=log sine As. Sine Gas=51° 32':sines=190.39':: rad: sine as=25o.26'. * 2. To find ag the hour from six, in the trianglè Ags. Rad x sine ag=tang Gs x cot GAS. (log tang Gs+log cot gas) -10=log sine ag. or, Rad : tang GS = 19°.39'::cot GAS = 51° 32' : sine ag = 160.28'.48". Now 169.28.48"=1h.5.55" the time from six, hence the sun will be east at 7h.5'.55" in the morning, or west at 4.6.54'.5'' in the afternoon. (E) As the declination increases, the altitude and hour from six increase, and when the declination vanishes, the sun appears in the horizon upon the prime vertical. When the latitude is equal to 90°, the sun's altitude upon the prime verticalis equal to his declination. When the latitude of the place is less than the declination, the sun never appears on the prime vertical. EXAMPLE II. At what time will Arcturus appear upon the prime vertical, or be due east or west from Greenwich, latitude 51°.28'.40" N., on the first of April 1822; and what will be its altitude ? Its right ascension being 146.7.32", and declination = 20°.6ʻ.50" North (Table VIII.), the sun's right ascension on the same day at noon (Naut. Alm.) being=05.41'.11", and on April 2d=0b.44'.49". 1. To find the altitude As, in the triangle AGS. Rad x sine gs=sine gas x sine As. :: (10+log sine Gs)—log sine gas=log sine As. or, Sine GAS=51°.28'.40" : rad::sine gs=20°.6'.50" : sine As = 26°.4'.31". * 2. To find Ag, the complement of the time from the star's cul minating, in the triangle AGS. Rad x sine Ag=tang Gs x cot GAS. (log tang Gs+log cot Gas) -10=log sine AG. or, Rad : tang ts=20°.6'.50"::cot GAS=51°.28.40? ; sine =160.57'. Hence 90°- 160.57' =730.3', which reduced to time = 46.52.12", the time from the star's culminating. The star comes to the meridian at 13h.24'.19" (D. 273.); therefore 13h.24'.19" – 45.52'.12" 8h.32'.7", the time in the evening when the star will appear due east; and 13h.24'.19" +4h.52.12 =18h.16.31"; or 65.16'.31" next morning, when the star will appear due west. (F) The height of the same star upon the prime vertical in any place, is always nearly the same, for the reasons already assigned (C. 290.); and the difference of the times of its coming to the prime vertical, will be equal to the difference of the times of its culminating, which is nearly equal to the diurnal difference of the sun's right ascension. PRACTICAL EXAMPLES. 1. The sun's declination being 19o.39' North, and having 25o.26' altitude upon the prime vertical; required the latitude of the place, and the hour of the day? Answer. Latitude - 510.32' N. time 75.5'.55" or 45.54'.5", according as the observation was made in the forenoon or afternoon. 2. At what time will Aldebaran appear due east or west at Greenwich, latitude 51o.28'.40' N. on the 20th of November 1822, and what will its altitude be at that time? P. Its right ascension being=4h.25'.43" and declination 169.8'. 36" N. (Tab. VIII.), the sun's right ascension at noon 20th November (Naut. Alm.)=156.41.29", and on the 21st 155.45'40". Answer. Aldebaran culminates at 125.42' (D. 273.), is due east at 7h.35'.17"; due west at 176.48.43', and its altitude 200.46'.8". The arc AG=130.19.19". 3. The sun's declination being 19o.39' N. and he was observed to be due west at 45.54' in the afternoon; required his altitude, and the latitude of the place? Answer. Altitude=259,26', latitude 510.32 N. 4. At what time will Regulus appear due east or west at Greenwich, latitude 519.28'.40' N., on the 6th of February 1822, and what will be its altitude at that time? Its right ascension being = 9h.58'.53", and declination 120.50.1 N. (Tab. VIII.) the right ascension of the sun at noon on the 6th of February (Naut. Alm.) being=216.18'.41", and on the 7th=315.22'41". Answer. Regulus culminates at 126.38.5", is due east at 75.19.56", due west at 17h56'.14", and its altitude=169.29'.37". The arc ag=10°.27.37". 5. In latitude 510.32 North, the sun's altitude when on the prime vertical was 25°26'; required his declination, and the hour of the day? Declination – 199.39' North, Answer. Time 76.5'.55", or 4h.54'.5", according as the observ, ation was made in the forenoon or afternoon. (Plate III. Fig. 1.) ni. (G) Given the latitude of the place, and the sun's altitude, when on the equinoctial, to find his azimuth and the hour of the day. In the right-angled spherical triangle adg. 1. Ag=the complement of Æg the hour from noon, being an arc of the equinoctial, on which g represents the sun's plave between noon and 6 o'clock. 2. Ad=the complement of hd the sun's azimuth from the meridian, or south point of the horizon. 3. dg=the sun's altitude. PROBLEM V. |