Example. Let the latitude be 50:48', and the declination of a celestial object 10:25? ; required the true amplitude of that object ? Latitude. 50:48? log. secant 10.199263 Declination 10.25 log. sine 9. 257211 True amplitude as required 16:37:22" log. sine . 9.456474 TABLE L. To find the Times of the Rising and Setting of a Celestial Object. This Table contains the semidiurnal arch, or the time of half the continuance of a celestial object above the horizon when its declination is of the same name with the latitude of the place of observation; or the time of half its continuance below the horizon when its declination and the latitude are of different denominations. The semi-diurnal arch expresses the time that a celestial object takes in ascending from the eastern horizon to the meridian; or of its descending from the meridian to the western horizon. As the Table is only extended to 23į degrees of declination, being the greatest declination of the sun, and to no more than 60 degrees of latitude ; therefore, when the declination of any other celestial object and the latitude of the place of observation exceed those limits, the semi-diurnal arch is to be computed by the following rule; viz., To the log. tangent of the latitude, add the log. tangent of the declination, and the sum, rejecting 10 in the index, will be the log. sine of an arch; which being converted into time, and added to 6 hours when the latitude and declination are of the same name ; or subtracted from 6 hours when these elements are of contrary names; the sum, or difference, will be the semi-diurnal arch. Example 1. Let the latitude be 61 degrees, north, and the declination of a celestial object 25:10:, north ; required the corresponding semi-diurnal arch? 619 0! north, log, tangent 10. 256248 . 25. 10 north, log. tangent 9.671963 Latitude Arch = 5757'21" = log. sine.. . 9.928211 Arch cony, into time nal arch, as required. 3:51" 49: + 6 = 9:51:49, the semidiur Example 2. Let the latitude be 20:40:, south, and the declination of a celestial object 30:29', north; required the corresponding semi-diurnal arch? Latitude. . 20:40: south, log. tangent . . 9.576576 Declination . 30. 29 north, log. tangent 9.769860 . . The present Table has been computed agreeably to the first example; but as in most nautical computations, it is not absolutely necessary that the semi-diurnal arch should be determined to a greater degree of accuracy than the nearest minute; the seconds have, therefore, been rejected, and the nearest minute retained accordingly. Since the Table for finding the time of the rising or setting of a celestial object (commonly called a Table of semi-diurnal and semi-nocturnal arcs,) is scarcely applied to any other purpose, by the generality of nautical persons, than that of merely finding the approximate time of the rising or setting of the sun ; the following problems are, therefore, given for the purpose of illustrating and simplifying the use of this Table; and of showing how it may be employed in determining the apparent times of the rising and setting of all the celestial objects whose declinations come within its limits. PROBLEM I. Given the Latitude and the Sun's Declination, to find the Time of its Rising or Setting RULE. Let the sun's declination, as given in the Nautical Almanac, be reduced to the meridian of the given place by Table XV., or by Problem I., page 76; then, Enter the Table with this reduced declination at top, or bottom, and the latitude in either of the side columns; under or over the former, and opposite to the latter, will be found the approximate time of the sun's setting when the latitude and declination are of the same name; or that of its rising when they are of contrary names.-The time of setting being taken from 12 hours will leave the time of rising, and vice versa, the time of rising being taken from 12 hours will leave that of setting. Note.- Proportion must be made, as usual, for the excess of the minutes of latitude and declination above the next less tabular arguments. Example 1. Nautical 21:49:51" north. 2'59" 7952: 4! 48! Tabular difference to l: of lat. = 4!; now + 3 60 3! x 17! Tab. difference to 30 of dec. = 3!; now =+ 2, nearly 30: Approximate time of the sun's setting 7.57 Approximate time of the sun's rising 4. 3 Note.-Twice the time of the sun's setting will give the length of the day; and twice the time of its rising will give the length of the night. Example 2. 3:16! 6" south, var. of dec. 23:20", and long. 105. E. 6:48" Sun's dec. reduced to the given meridian 3: 9:18", or 3:9: south. Time in Table L., ans. to lat. 40. north, and % 0' x 30 Tab, diff. to 10 of lat. =(! ; now 0 60 31 x 9 Tab. diff. to 1. of dec. = 3! ; now 0 60: dec. 3: south, is . 6.10 Approximate time of the sun's rising. 6.10 Approximate time of the sun's setting 5.50 The nearest minute of declination is sufficiently exact for the purpose of finding the approximate times of the risiog and setting of a celestial object. Remark. Since the times of the sun's rising and setting, found as above, will differ a few minutes from the observed, or apparent times in consequence of no notice having been taken of the combined effects of the horizontal refraction and the height of the observer's eye above the level of the sea, by which the time of rising of a celestial object is accelerated, and that of its setting retarded ; nor of the horizontal parallax which affects these times in a contrary manner; a correction, therefore, must be applied to the approximate times of rising and setting, in order to reduce them to the apparent times. This correction may be computed by the following rule; by which the apparent times of the sun's rising and setting will be always found to within a few seconds of the truth. Rule.-To the approximate times of rising and setting, let the longitude, in time, be applied by addition or subtraction, according as it is west or east, and the corresponding times at Greenwich will be obtained : to these times, respectively, let the sun's declination be reduced by Table XV., or by Problem I., page 76, then, Find the sum and the difference of the natural sine of the latitude, and the natural co-sine of the declination (rejecting the two right hand figures from each term), and take out the common log. answering thereto, rejecting also the two right hand figures from each :-now, to half the sum of these two logs. add the proportional log. of the sum of the horizontal refraction and the dip of the horizon diminished by the sun's horizontal parallax, and the constant log. 1. 1761* ; the sum of these three logs., abating 4 in the index, will be the proportional log. of a correction ; which being subtracted from the approximate time of rising, and added to that of setting, the apparent times of the sun's rising and setting will be obtained. Thus,-Let it be required to reduce the approximate times of the sun's rising and setting, as found in the last Example, to the respective apparent times ; the horizontal refraction being 33!; the dip of the horizon 5:15", and the sun's horizontal parallax 9 seconds. The sun's declination reduced to the approximate time of rising, is 3:3:37", and to that of setting 3:14:58" south. * This is the proportional log. of 12 hours esteemed as minutes. Note. In this method of reducing the approximate to the apparent time of rising or setting, it is immaterial whether the latitude and declination be of the same, or of contrary names : -nor is it of any consequence whether the declination be reduced to the approximate times of rising and setting or not, since the declination at noon will be always sufficiently exact to determine the correction within two or three seconds of the truth, On account of its natural co-sine being only required to four places of figures :--this will appear evident by referring to the above example, where, |