5. Again, in the triangle znm, we have zm=90° +33'— 56'.59" — 890.36'.1" To find the Given Nm=90° +8°.38.36" = 989.38.36" LZNM. ZN=90°-51°.28.40" = 380.316.20" The L znm will be found=789.20'.4" = 51.13.20", the time the moon rises before she comes to the meridian; whence 13h.32'.51' – 5h.13'.20"=8h.19'.31"P.M. thetime of the moon's rising, nearly. To find the correct time. The ) passes the meridian on the 5th of October (Prob. viii. page 276.) at 145.18'.12", then 14h.18'.12" - 13.32.51"=45.21" daily variation of transit; hence, 24" : 45'.21"::8h.19.31" : 15'.44", and 8h.19'.31" – 15'.43"=8h.3'.47" P.M. true time of the moon's rising. (T) Should more examples be thought necessary, they may be formed and solved in a similar manner; likewise the rising of the different planets may be found by the same method. The times of the rising and setting of the moon and of the planets, are given in White's Ephemeris, and in the Connaissance des Tems. These works will serve as a check upon the calculation. The rising and setting of the moon, and of the planets, are not inserted in the Nautical Almanac. PROBLEM X. (Plate III. Fig. 1.) (U) The latitude and longitude of a fixed star, or planet, being given, to find its right ascension and declination, et contra. * In the oblique triangle nnc. Nn=obliquity of the ecliptic, or distance between the poles of the equator and ecliptic. nc=the complement of the star's latitude vc. Nc=the complement of the star's declination RC. The angle nnc, measured by the arc v®, is the complement of the star's longitude from aries. The supplement of the angle nnc, measured by the arc'RQ, is the complement of the star's right ascension from aries. Or, In the oblique triangle smx. * A neat method of solving this problem may be seen at page 44 and 45, Vol. I. third edition, of Dr. Mackay's useful treatise on the theory and practice of finding the longitude, or in Dr. Maskelyne's introduction to Taylor's logarithms. The problem may be varied so as to admit of several cases, but the two above specified are the only useful ones. The supplement of the angle xms, measured by the arc vv s, is the complement of the star's longitude from aries. The angle xsm, measured by the arc rg, is the complement of the right ascension from aries. EXAMPLE. The latitude of Aldebaran in Taurus being=50.28'.27" S.; its longitude 29.69.56:24"*; required its right ascension and declination; the obliquity of the ecliptic being 230.28'? In the triangle smx. 1. To find the co-declination sx. 38053 mx=840.31'.33" log sine = 9.99802 {xms=780.28.12"log sine= {9:99114 76106 mx-sm=610.3'. 33/ Nat. cos= 48891 22o.22' log sine Nat. cos 73o.541.37" 27715 sms 2 9.58042 SX=1060.5'.23" obtuse. 2. To find the co-right ascension xsm. Sine mx=84o.31.33" : sine xsm=230.56'.22". OR THUS: Draw the perpendicular XL. PRACTICAL EXAMPLES. 1. Required the right ascension and declination of Regulus in the Lion's heart; latitude being 0°.27'.30” N., and longitude 48.260.57'.5", obliquity of the ecliptic being 230.28'? Answer. Right ascension=149o.19'.54" and declination= 120.58'.21" N. 2. Required the right ascension and declination of Spica Virginis, in the sheaf of Virgo; latitude 20.2.23" South, and longitude 68.209.571.10"; obliquity of the ecliptic 230.28'? The latitudes and longitudes of the stars, in this, and many of the following examples, in the first edition of this work (published in the year 1801) were adapted to the year 1796; any change of data is at present unnecessary, as the method of solution remains the same; being independent of an ephemeris, and the answers, when obtained, would only differ a few seconds from those now given. Answer. Right ascension=1980.34.32", and declination= 100.4'.31" South. 3. Required theright ascension and declination of Betelguese in the shoulder of Orion; latitude 160.3.3" South, longitude 25.250.51'.46"; obliquity of the ecliptic 23o.28? Answer. Right ascension=859.59'.28", declination=70.21'. 17' North. 4. Required the right ascension and declination of Antares in Scorpio; latitude 40.32.35" S., longitude 89.69.51.51"; obliquity of the ecliptic 230.28' ? Answer. Right ascen.=244°.11'.3", declination=259.57.22" South. 5. Required the right ascension and declination of the moon, when her latitude is 40.0.34" N., longitude 78.149.26'.21", and the obliquity of the ecliptic 23°.27'.48" ? Answer. Right ascension=2230.11'.11", declin.=12o.21.14" South. 6. Given the right ascension of Aldebaran=46.25'.43", and its declination 1608'.36'' N. in the year 1822, required its latitude and longitude ; the obliquity of the ecliptic being 23° 28' ? In the triangle nNC. 1. To find the co-latitude nc. NN=230.28' log sine 9•60012 || Nat. sine 210.30ʻ.12" =36655 Nc=790.51'.24" log sine ' func=789.12.52" log sine = 9.99075 NC-Nn=50°. 23.24' Nat. cos = 21°.30.12" log sine= 9.56414 Nat. cos 84o.31'.1" 9.98252 9.99075 73310 =63755 09555 Co-lat. nc=950.28'.59" obtuse. Hence the latitude=50.28'.59" South. OR THUS, Draw the perpendicular cl, Then, in || In the oblique-angled triangle nnc the right-angled triangle CLN, (B. 178.) rad, sine of 90° 10.00000 COS NL=72o.28'.25" 9.47877 : COS CNL=RQ=230.84'. 15" 9.96216 : COS NL=950.56'.25" 9.01490 stang Ne=3.51.24 10.53842 :: COS NC = =730.51'.24" 9.44417 : tang NL=720.281.25". 9.50058 : COS nc= =840.30'.591 8.98030 Then NL + NN=NL=950,56.25" Hence the lat=59.29'.1" South. * The latitude of this star, on the same day, in the year 1796, was 50.28',27" S. making a difference of only a few seconds in 26 years. (See the note page 306.) 2. To find the co-longitude cnn. 7. Required the latitude and longitude of Sirius in the Great Answer. Latitude 39o.33.30% S. longitude 35.11o.13'.44". 8. Required the latitude and longitude of Procyon in the Little Dog; right ascension being 1120.6'.47", declination 5o. 45'.3", N.; obliquity of the ecliptic 239.28'? Answer. Latitude 150.58.14" S., longitude 35.22 .55'.42". 9. Required the latitude and longitude of Arcturus in Boötes: right ascension 2119.33.17", declination 20°.15.58" N.; the obliquity of the ecliptic being 23o.28? Answer. Lat. 30°.52'.43" N., long. 68.219.20.41". 10. Required the latitude and longitude of Fomalhaut, in the Tidl. Answer. Lat. 5°.6' N., longitude 45°.28'. PROBLEM XI.' (Plate III. Fig. 8.) Required the distance between Sirius in the Great Dog, * The longitude of this star in the year 1796 was found, by the same process, to be 660.56' 24", making an increase of 214.43" in 26 years. By comparing together the places of the fixed stars deduced from observation, astronomers have found that their longitudes increase about 504" annually. This increase of longitude must necessarily cause an irregular motion in the same star with regard to the equinoctial. Hence the right ascensions and declinations of stars are continually varying, so that stars which formerly had north declination have now south declination, et contra. Their latitudes are also subject to a small variation, See the Tables, page 153, &c. Naut. Alm. 1822. In the triangle xst let the star at t represent Procyon, because its right ascension is the greatest, and the star at x represent Sirius. The 2 xst=139.6'.26" diff. of their right ascensions-RQ. sx=730.33.25" the complement of Sirius's declination rx. ST=95°.45'.3" Procyon's declination +90°=5Q+QT. Hence, here are two sides and the contained angle given, to find the third side xt, by any of the methods in the preceding problems. Answer. xt=250.42'.10'. EXAMPLE II. The latitude and longitude of Sirius being=39o.33.30”Si, and 3.119.13'.44"; and the latitude and longitude of Procyon =159.58'.14" South, and 35.220.55'.42". Required their distance ? In the triangle xmt as before, we have xm=50°.26'.30" the complement of Sirius's lat. vx. tm=74o.1'.46' the complement of Procyon's lat. & T. The _xmt=119.41.58" the difference of their longitudes Hence, as before, here are two sides and the contained angle given, to find the third side, xt=25o.42 Answer. EXAMPLE III. Required the distance between Capella in the Goat, right ascension=75°.21'.19", declination=45°,46'.15' North, and Procyon in the Little Dog, right ascension=1120,6'.47", and declination=59.45'.3'' North? In the triangle TNC, let the star at T represent Procyon, and ç Capella. NC=44°.13.45% the co-declination of Capella. TN=849.14'.57" the co-declination of Procyon. LTNC=360.45'.28" the diff. of their right ascensions=RQ. Here we have two sides and the contained angle given, to find the third side, as in the preceding example. Answer. CT=51°.6'.56". EXAMPLE IV. Required the distance between Capella, whose latitude is 229.51'.57" North, longitude 28.189.57'.5.7", and Procyon, latitude 150.58'.14" South, longitude 35.220.55.42"? |