(X) This solution is analogous to Problem 8th in the Requisite Tables published by order of the Commissioners of longitude; or to Problem VII. Chap. II. Book III. of Dr. Mackay's Navigation. If you have the Requisite Tables* by you, the sum of the three logarithms above may be looked for in the column of rising, and the hours and minutes answering thereto will be the time from noon. Or, increase the index, by 5, and the sum is the log., versed sine of the same angle. Or, the log. versed sine of an arc increased by the logarithm of 5 (with 9 for its index), will give double the logarithmical sine of half the required angle. (P. 125.) EXAMPLE II. In latitude 39°.54' North, longitude 80°.39'.45" West of Greenwich, suppose the altitude of the sun's lower limb to be 15°.40′.57′′ on the 8th of May 1813, at 5h.30′.32′′ P. M. per watch. Required the error of the watch. 1. To find the time at Greenwich, nearly. 80°.39′.45′′ × 4=51.22′.39′′ the difference of time between the clocks at Greenwich and the place of observation. Hence, 5.30′.32"+5.22′.39"=10h.53′.11" P.M. thetime at Greenwich. 2. To correct the O's declin. ||3. To correct the O's altitude. O's decl. May 8th 1813 =17°. 3'.22''s observed alt. Ditto May 9th =15°.40′57′′ =- 3. 17 Increase in 24 hours 15. 37.40 The XVIth of the Requisite Tables, or the XXXV, XXXVI, and XXXVII Tables in Dr. Mackay's Navigation. The XVIth of the Requisite Tables is divided into three columns. The FIRST column, called half elapsed time, is only a table of secants with 10 rejected from the index; the time being reduced to degrees. The SECOND, called middle time, is the logarithmical sine of half the elapsed time, increased by the logarithm of 2, and the index of the sum diminished by 5. The THIRD, called column of rising, is a table of logarithmical versed sines with their indices diminished by 5. Thus for 1 hour of time, or 15 degrees. co-sec of 15°10'58700, sine of 15° log. of 2 = 941300, ver. sine of 15°-8-55243 =0.30103, -5 Proceed in the same manner for any other time. + Table IV. Nautical Almanac. Table VI. 4. To find the true time. In the triangle szy. Complement of latitude sum Complement of altitude NZ=50°. 6'. 0" cosecant sz=74. 6.19 SN 72. 49.18 2 197. 1.37 reject. indices. 11511 cosecant ⚫01982 98. 30.48 sine (Y) In the practical application of this problem, it will be proper to take several altitudes of the sun within a minute or two of each other, and to note the corresponding times per watch. The mean of these altitudes, and of the times, will be more correct than a single altitude and time, and will in a great measure counteract the errors arising from the imperfection of the instrument. EXAMPLE III. On the 4th of March 1313, in latitude 45°.37′ North, and longitude 93.41'.30" West, suppose the following altitudes of the sun's lower limb to be observed: Required the apparent time of observation, and the error of the watch. 002 Answer. Answer. The apparent time of observation 2h.55′.47", watch too slow 41′′. EXAMPLE IV. On the 16th of September 1813, in latitude 33°.56' South, and longitude 25°.9' West, suppose the mean time per watch of several altitudes of the sun's lower limb to be 8h.12.10" A.M., and the mean of the altitudes themselves 24°.43'; required the error of the watch? Answer. The apparent time is 3.49.4" from noon, or 8h.10'.56" A. M., hence the watch is 1'.14" too fast. (Z) Given the latitude of the place, the declination and altitude of a known fixed star, to find the hour of the night when the observation was made. EXAMPLE I. In latitude 7.45' South, and longitude 30°.18' East of Greenwich, on the 7th of September 1813, suppose the altitude of the star Procyon, when east of the meridian, to be 28°.12'; required the true time of the day? (A) The hour 4 found by calculation, as above, is the distance of the star from the meridian of the place of observation; consequently when the star is eastward of the meridian, at the time of observation, its right ascension must be diminished by the hour angle, and when it is westward of the meridian its right ascension must be increased by the hour angle; and the the difference, or sum, will be the right ascension of the ob server's meridian. From the right ascension of the observer's meridian (increased by 24 hours if necessary) subtract the sun's right ascension, as given in the Nautical Almanac, for the noon preceding the observation, the remainder will be the approximate time at the place of observation. To this time apply the longitude of the place of observation, by addition or subtraction according as it is westward or eastward, and the result will give the time at Greenwich nearly. Calculate the increase of the sun's right ascension for the time at Greenwich, and subtract it from the time at the place of observation, found above, the remainder will be the correct apparent time required, (B) In order to attain the greatest accuracy several altitudes of a star should be observed in succession, and if two or more separate stars be observed, and the error of the watch be deduced from each star, the medium result will be more correct than that derived from the observation of a single star. If an equal number of stars can be observed on each side of the meridian, and nearly equidistant therefrom, the errors arising from the imperfection of the instrument, &c. will be rendered almost insensible. EXAMPLE 11. On the 29th of January 1813, in latitude 53°.24' North, and longitude 25°.18' West; suppose the following altitudes of Procyon to the west, and Alphacca to the east of the meridian, were observed by two separate persons at the same instant of time required the error of the watch. The right ascension of Alphacca being 15.26′46′′ and declination 27°.24.51′ N. Time Answer. The hour by Alphacca's altitude is 3.41′.15′′, and the correct time 14h.55'.46"; hence the watch is 2'.52" too fast. The hour by Procyon's altitude is 4h.19'.35", and the correct time 14h.59'.19"; hence the watch is 50" too fast. EXAMPLE III. In latitude 48°.56' North, longitude 66° west, suppose the altitude of Aldebaran, when west of the meridian to be 22°.20.15" on the 15th of April 1813; required the apparent time of observation? Answer. The hour =4h.57.28", and the correct apparent time 7.47'.45". (C) SCHOLIUM. The three preceding problems are all included in one triangle, but they have been considered separately, on account of their importance, and to prevent confusion. Any three of the five quantities, viz. the sun's or star's declination, the latitude of the place, the altitude of the object, the azimuth, and the hour of the day being given, the rest may be found. A sixth quantity might be taken into consideration, were it materially useful, viz. the opposite to the complement of the latitude, being formed by the intersection of the vertical circle zsc, and the azimuth circle NSD. PRACTICAL EXAMPLES. 1. Given the sun's declination 19°.39' North, in North latitude,his corrected altitude 38°.19',and his azimuth=S. 72°.13′E.; required the hour of the day and the latitude of the place? Answer. The hour angle is 52°.30'=3h.30′, the hour from noon, hence the time is 8h.30' A. M., and the latitude 51°32′ North. When the altitude of the sun and his declination are equal, the triangle are isosceles, and the hour angle is equal to the azi muth. |