12th of October 1813; the sun's declination at noon, for the meridian of Greenwich, being 7°.21'.18" S*. Answer. The sun's declination, when it is noon at Glasgow, is 7°.21.34" S. The O's declination being south the triangle ZNO will fall on the left hand of the prime vertical za. Then zO=90°.32.51 NO 97°.21'.34" ZN=34°.28′ With these three sides find the hour 4ZNO 80°.8′.52′′=5".20'.35′′the time from noon when the sun sets, hence the sun rises at 6".39′.25′′. Call these the approximating times of rising and setting; calculate the sun's declination to the approximating time of rising and setting, which you will find to be 70.16′.32′′S,and 7°.26′.35′′S; then, with the polar distance NO=97°.16′.32" repeat the operation, and the true apparent time of rising will be 6". 38'.54"; with the polar distance NO=97°.26′.35′′ repeat the operation a second time, and the true apparent time of the sun's setting will be 5.20.5". To obtain the true, or mean, time of rising and setting, as shewn by a well regulated clock, or time-piece, the equation of time must be applied as directed in page II, and page 149 of the Nautical Almanac. PRACTICAL EXAMPLES. 1. Required the apparent time of the rising and setting of the sun, in latitude 51°.32′ N. supposing his declination to be 23.29' S. Answer. 8.8'.2". and 3",51'.58", 2. Required the apparent time of the rising and setting of the sun, in latitude 51°.32′ N. when his declination is 170.32′ N. supposing it to undergo no change from sun-rise to sun-set. Answer. 4.22′.16′′ and 7.37′.44′′. To reduce the sun's declination, as given in the Nautical Almanac, to any other me ridian, and to any given time of the day. RULE. Turn the given longitude into time (see the note page 247) and add it to the time at Greenwich if it be west longitude, or subtract it if east. Find the change of the sun's declination in 24 hours from the Nautical Almanac ; Then say, as 24 hours: to this change: : the given time from noon: the variation of the sun's declination in that time. This variation must be added to the sun's declination at noon, or subtracted' from it, according as the declination is increasing or decreasing. + Long, of Glasgow=4°.15′-17m of time, hence when it is noon at Glasgow it is 0.17 at Greenwich. O's declination at noon October 12th, is Increase of declination in 24 hours 70.21'.18" 70.45.51" 0°.22'.33" Then 24h: 22.33"; - 17m: 16′′ the increase of the sun's declination in 17 minutes; consequently when it is noon at Glasgow the sun's declination is 70.21′.34′′ S. M m 2 3. Required 3. Required the apparent time of the rising and setting of the sun in latitude 56° N., supposing the sun's declination at his rising to be 22°.34.35", and at his setting 22°. 30′.7". Answer. 3.22′.20" and 8".37′. PROBLEM IX. (Plate III. Fig. 7.) (H) Given the latitude of the place, the day of the month, the moon's horizontal parallax and refraction, to find the time of her rising and setting. Required the time of the rising and setting of the moon at Greenwich, latitude 51°.28'.40" N., on the 13th of August 1813. The horizontal parallax of the moon being from 53′ to 62′, always exceeds the horizontal refraction; therefore, when the moon's upper limb appears in the horizon she is really above it, by a quantity equal to the horizontal parallax minus the refraction. The moon's declination must be found as near to the time of her rising as possible, because it is subject to a considerable variation in the course of a few hours. Let n be the place of the moon, when in the horizon, m the point where she becomes visible, and e her place on the meridian; then de will represent her change of declination from the time of her rising to the time of her transit. Find the time of the moon's passage over the meridian *, and calculate the declination to that time (see the Note page 267). With this declination, the latitude, &c. calculate the hour angle, as in the preceding problem, and call this the estimated time of the moon's rising; correct the moon's declination, and her horizontal parallax, to the estimated time of rising, and repeat the operation till the time of rising be accurately ascertained. *To find the time of the moon, or any planet's culminating." Take the difference between the sun's and planet's motion in right ascension in 24 hours, if the planet be progressive, or their sum, if retrograde. Then, As 24 hours diminished by this sum or difference, when the planet's motion is greater than the sun's, or increased by it when the sun's apparent motion is greater, is to 24 hours; so is the planet's right ascension at noon, diminished by the sun's, to the time of its transit. If the sun's right ascension be greater than the planet's, 24 hours must be added to the planet's right ascension before you subtract. Note. The right ascensions of the planets are not given in the Nautical Al- · manac; but they may be calculated from their geocentric latitudes and longitudes, which are inserted in that work, and the obliquity of the ecliptic (see the succeeding problem.). SOLUTION. SOLUTION. gh.34.27" 1. To find the time of the moon's passage over the meridian. Oright ascen. Aug. 13, at noon † 9.30′41′′ D's right ascen. at Ditto, 339°.52' Decrease of motion in 24 hours 30.46" Decreases in 24 hours 13°.6 13°.6' x 4 52.24"; and 52.24"-3".46" 48".38" the moon's motion exceeds the sun's in 24 hours. D's right ascension 339°.52′ x 422.39'.28", which di minished by the O's right ascension, leaves 13.8.47". Then 24-48.38": 24h:: 13.8.47" : 13.36.21" the true time of the moon's passing the meridian at Greenwich. 2. To find the moon's declination, when she is on the meridian. D's declination at midnight, 13th August 1813§=8°.59'.S. D's declination at noon, 14th August 1813= Decrease of declination in 12 hours= 6°.46'.S. 2°.13'. 12h.: 2°.13′ :: 1.36′.21" : 17'.47" the change of the moon's declination from midnight, on the 13th of August, to the time of her transit; hence 80.59'-17.47" 8°.41.13" the moon's declination, when on the meridian. 3. In the triangle zNm, find the ▲zNm. = The refraction =33', the D's horizontal parallax, from the Nautical Almanac, 57.55" at midnight, and by calculating the proportional part for 1".36′.21", the time past midnight when the moon is on the meridian, the horizontal parallax at that time will be 57.57". Hence, we have `zm=90° +33′-57′.57′′-89°.35'.3" Given Nm 90°+ 8°.41'.13′′=98°.41′.13" ZN=90° 51°.28.40"-38°.31'.20" To find the LZNM. The 4zNm will be found=78°.14.44" 5.12'.59"=nearly the time the moon rises before she comes to the meridian. Whence 13.36.21"-5.12.59"-8".23′.22" the estimated time of the moon's rising. 4. To find the D's declination at the estimated time of rising. + Nautical Almanac. The time of the moon's passing the meridian of Greenwich is given, in the VIth page of each month, in the Nautical Almanac. § Nautical Almanac. D's D's decl. at noon 13th August 1813 11°3' S. | | Horizontal parallax= 57.40" 80.59 Horizontal parallax= 57.55′′ = D's decl. at midnight ditto 5. Again, in the triangle zNm, we have zm=90°33′-57′.50"-89°.35′.10′′ Given Nm 90° + 9°.56′19′′-99°.36′.19′′ ZN 90° 51°.28'.40"-38°.31'.20" To find the LZNM. The 4zNm will be found=77°.2′10′′=5.8'.9" the time the moon rises before she comes to the meridian; whence 13".36′.21′′ - 5.8′.9′′=8.28′.12" the time of the moon's rising, nearly. If greater exactness be required, find the moon's declination at 8.28′. 12′′ and repeat the operation. (I) Should more examples be thought necessary, they may be formed and solved in a similar manner; likewise the rising and setting 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 Connoissance des Temps. 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.). (K) 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 NC, measured by the arc RQ, is the complement of the star's right ascension from aries. Or, In the oblique triangle smx. sm the obliquity of the ecliptic. mx the complement of the star's latitude vx. * 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 prac tice of finding the longitude The The supplement of the angle xms, measured by the arc vv , is the complement of the star's longitude from aries. The angle xsm, measured by the arc RQ, is the complement of the right ascension from aries. EXAMPLE. The latitude of Aldebaran in Taurus being50.28′.27′′. S., its longitude 2.6°.56'.24′′*; required its right ascension and declination; the obliquity of the ecliptic being 23°.28'? In the triangle smx. sm = 23°.28′. 0"the obliquity of the ecliptic. mx = 84°.31′.33" the complement of the star's lat. 1. To find the co-declination sx. sm=230.28'.00" log sine= 9.60012 Nat. sine 22°.22′ = - 38053 Hence 106°.5′.23′′ — 90°—16°.5′.23′′—RX the declination N. 2. To find the co-right ascension xsm. Sine sx=106°.5′.23" : sine xms=156° 56′.24" :: Sine mx=84°.31′.33" : sine xsm=23°56′.32". Hence AR 66°.3′.28", the right ascension. OR THUS: Draw the perpendicular XL. Rad: cos Lmx:: tang mx : tang Lm, and Lm+ms=LS. Sine mL sine SL :: cot Lmx: cot xsm. PRACTICAL EXAMPLES. 1. Required the right ascension and declination of Regulus in the Lion's heart; latitude being 0°.27′.30" N., and longitude 4.26°.57.5", obliquity of the ecliptic being 23°.28/? Answer. Right ascension=149°.19.54" and declination= 12.58 .21" N. 2. Required the right ascension and declination of Spica Virginis, in the sheaf of Virgo; latitude 2°.2'.23" South, and longitude 6.20°.57.10"; obliquity of the ecliptic 23°.28'? * The latitudes and longitudes of the stars, in this, and many of the follow ing examples, were calculated for the year 1796; they have not been altered in this edition, because the solutions do not depend on an Ephemeris, and the answers would only differ a few seconds from those now given. Answer |