(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 rwill be due east and west. 1. AG In the right-angled spherical triangle ags. 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 HE 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 51°.32' North, when the sun's declination is 19°.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. or, SineGAS 51°.32′: sine Gs=19°.39′:: rad: sine As=25°.26'. 2. To find AG the hour from six, in the triangle AGS. * Rad x sine AG=tang GS × 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 16°.28'.48" 1.5'.55" the time from six, hence the sun will be east at 7.5′.55′′ in the morning, or west at 41.54′.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 14.7.32", and declination, ≈ 20°.6′.50′′ North (Table VIII.), the sun's right ascension on the same day at noon (Naut. Alm.) being=0.41'.11", and on April 2d 0.44'.49". 1. To find the altitude as, in the triangle AGS. * Rad x sine GS=sine GAS X 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 culminating, 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 GS=20°.6'.50":: cot GAS-51°.28'.40" sine AG =16°.57'. Hence 90°-16°.57′=73°.3′, which reduced to time= 4h.52.12", the time from the star's culminating. The star comes to the meridian at 19h.24.19" (D. 273.); therefore 13h.24.19′′ - 4h.52′.12′′=8h.32.7", the time in the evening when the star will appear due east; and 13.24′.19′′+4.52.12′′ 18.16.31"; or 6h.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 19°.39′ North, and having 25°.26' altitude upon the prime vertical; required the latitude of the place, and the hour of the day? Answer. Latitude 51°.32′ N. time 7.5'.55" or 4h.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 51°.28.40" N. on the 20th of November 1822, and what will its altitude be at that time? Its right ascension being 4.25'.43" and declination 16°.8'. 36" N. (Tab. VIII.), the sun's right ascension at noon 20th November (Naut. Alm.) 15.41'.29", and on the 21st 15.45.40". Answer. Aldebaran culminates at 12.42' (D. 273.), is due east at 7h.35.17"; due west at 17h.48'.43", and its altitude= 20°.46'.3′′. The arc AG-13°.19.19". 3. The sun's declination being 19°.39' N. and he was observed to be due west at 4h.54' in the afternoon; required his altitude, and the latitude of the place? Answer. Altitude -25°.26', latitude-51°.32′ N. 4. At what time will Regulus appear due east or west at Greenwich, latitude 51°.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 12°.50.1" N. (Tab. VIII.) the right ascension of the sun at noon on the 6th of February (Naut. Alm.) being=211.18′.41′′, and on the 7th-21h.22.41′′. Answer. Regulus culminates at 12b.38'.5", is due east at 7h.19'.56", due west at 17h.56'.14", and its altitude=16°29′.37′′. The arc AG-10°.27′.37′′. 5. In latitude 51°.32′ North, the sun's altitude when on the prime vertical was 25°.26′; required his declination, and the hour of the day? Declination19°.39′ North. Answer. Time 7.5.55", or 4h.54.5", according as the observation was made in the forenoon or afternoon. PROBLEM V. (Plate III. Fig. 1.) (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 EG the hour from noon, being an arc of the equinoctial, on which & represents the sun's place between noon and 6 o'clock. 2. Ad the complement of нd the sun's azimuth from the meridian, or south point of the horizon. 3. do the sun's altitude. 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 be found. This triangle is complemental to the triangle GÆ zenith; for Gz is the complement of de the altitude; EG is the hour from noon; Æz the latitude; and the angle ÆzG, measured by the arc нd, is the sun's azimuth from the meridian. EXAMPLE. In latitude 51°.32' North, when the sun was on the equinoctial, his altitude was 22°.15'; required his azimuth, and the hour of the day? 1. To find ad the complement of the azimuth, in the triangle. AdG. * ** (log tang de+log cot Gad)—10=log sine ad. 38°.28′ : sine ad = Hence the azimuth-59°.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 ado. *. Rad x sine desine AG X sine Gad. (10+log sine dG)-log sine GAd=log sine AG. or, Sine Gad=38°.28': rad::sine de 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. PRACTICAL EXAMPLES. 1. In latitude 51°.32' N. when the sun has no declination, what is his altitude, and azimuth at three hours and a half from noon? Answer. The altitude 22°.15'. Azimuth S. 59°.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 8h.30'. Latitude 51°.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=57°46'; to find the hour of the day and the altitude and azimuth of the sun?, Answer. Time 8h.29'.56", or 3h.30.4". Altitude 22°.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 wooE (supposing wooE to be a part of the equator) than it is on the parallel of latitude MN, and greater at MN than at lL, &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, oo, &c. are the measures of the spherical angles WPO, OPO, &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? Answer. 3600 miles. |