It may frequently happen that the objects are too near the meridian to deduce the true time at the place of observation from either of the altitudes with required exactness; or, though the altitudes may be exact enough to use in clearing the distance, they may not be sufficiently so for deducing the time from them. In either case the error of the watch must be found from some other observation, and this error, being applied to the time at which the lunar distance is taken, will give the time at the meridian at which the observation for the error of the watch was taken, and the longitude thence deduced will be the longitude not of the place at which the distance is observed, but of that for and at which the error of the watch was found. In this manner may the results of a great many lunar observations be all referred to one place, and the situation of the ship at that time determined with a certainty to which the result of one observation can have no claim. The situation of the ship being thus determined by lunars, her longitude may be kept by the chronometer, till another opportunity is afforded to determine her place with precision by another series of lunars. The latitude used in computing the time is understood to be obtained from the course and distance run since the last observation for the latitude. An error of a few degrees in the longitude by account, or of half an hour in the time by account can be of very little importance, as the Greenwich time by account, which they are used in finding, is only employed in taking out the semidiameter and horizontal parallax of the moon from the Nautical Almanac, and these vary in general but a few seconds in twelve hours; the parallax sometimes twenty-four seconds; and this is about its maximum variation in twelve hours. If the time is not computed from either of the altitudes taken with the distances, the time of observation by a watch must be carefully noted, and the error of that watch found from some other observation; but, if the time is computed from one of the altitudes used in the lunar, no great care is required in noting the time. When speaking of altitudes and distances, it must always be understood that the means of several altitudes and distances are meant, when it is possible to obtain them. An observer may himself take both the altitudes and distances, first taking an altitude of each object, then a series of distances, and again the altitude of each object, carefully noting the times, and then finding by proportion what the altitudes must have been at the time of the mean distance. It may however frequently happen that the distances may be observed when from darkness or fog the horizon cannot be seen, and consequently the altitudes cannot be obtained. The altitudes must then be computed, and the method of computing them is shown in the following problem. The latitude of a place and the time being known, and the longitude by account, to compute the altitude of any known celestial object. If the object be the sun, the apparent time is the meridian distance, if less than twelve hours, but if it is more than twelve hours its compliment to twenty-four hours is the meridian distance. For any other celestial object, to the apparent time add the sun's right ascension, reduced to the given instant of Greenwich time, and from the sum subtract the object's right ascension, and the remainder will be the meridian distance of the object. If it exceed twelve hours take its complement to twenty-four hours as the meridian distance. With the latitude of the place, the polar distance of the object, and its meridian distance in degrees, proceed to compute the altitude as follows: Add together the cosine of the meridian distance and the cotangent of the latitude, and the sum rejecting ten from the index will be the tangent of arc first. If the meridian distance is greater than 90° take the sum of arc first and the polar distance, otherwise their difference, for arc second. Add together the secant of arc first, the cosine of arc second, and the sine of the latitude, and the sum rejecting the tens from the index will be the sine of the true altitude. If the apparent altitude is required, take the correction corresponding to the true altitude, and apply it with a contrary sign to the true latitude, and the result will be the approximate apparent altitude. From this approximate apparent altitude take out the correction, and, applying it with a contrary sign to the computed true altitude, the corrected apparent altitude will be obtained. In fig. 9, plate II., if A, S, and r, repre- otherwise their sum is CA; and cos. PC: = Example 1.-What are the true and apparent altitudes of the sun October 3d, 1828, at 3h. 4m. 12s. mean time, in lat. 40° 12′ N., long. by ac count 18° W. App.alt. 27 5 43 Example 2.-On March 9th, 1828, in lat. 36° 27' N., long. 35° W., at 7h. 2m. 10s. P. M., mean time, required the true and apparent altitudes of Procyon? h. m. s. 2.10 10 38 The sun's R. A. at 9h. 11m. 32s. is 23h. 21' 7s., the 's R. A. 7h. 30m 19s. and polar dist. 84° 21′. 7 app. time O's R. A. In computing altitudes it is absolutely necessary that the apparent time for the meridian of the place of observation should be known; therefore, when the error of the watch is found for some other meridian, the difference of longitude made from that meridian to that of the place of observation must be applied to the time shown by the watch, adding it if east, but subtracting it if west, to obtain the time at the place of observation. Let the time shown by the watch, e its error as found by observation, the difference of longitude made subsequently, and T the true time. Then Tel, l' being reduced to time, and e+when the watch is slow and when fast, and when the difference of longitude is east, and when west. 54 25 41 If the error was in mean time, this must be reduced to apparent time by applying the equation of time with a contrary sign. To find the variation of the compass.-If the bearing of any object by the compass be compared with its known bearing, the variation or deviation of the points of the compass from their corresponding points in the horizon becomes of course immediately known; the difference of the true and observed bearing being the variation. Now when an object is on the meridian its true bearing is either due north or due south; hence the deviation of an object from the meridian as observed by the compass, when the object is known to be on the meridian, is the variation of the compass; west when the object bears to the right; and east when it bears to the left of the meridian. Again, the middle time between equal altitudes of a celestial object being the time of its being on the meridian, the middle point between those on which it bears when it has equal' altitudes is its bearing when on the meridian; hence if this middle point be to the right of the meridian the variation is west; if to the left the variation is east. Example 1.-The sun at noon was observed to bear S. by W. W., what was the variation? S. by W. W. is 14 point to the right of S. therefore the variation is 14 point W. Example 2.-The sun, when he had equal altitudes on the same day, bore N. N. E. and N.W. by W.; what was the variation? The middle point between N. N. E. and N.W. by W. is N. by W. W., the bearing of the sun when on the meridian, which being 1 point to the left of N., the variation is 14 point east. The variation of the compass may also be found by the amplitudes of celestial objects. But, as from the effect of refraction they appear in the horizon when they are about 33′ below it, the centre of the sun, or a star, ought to be about 33′ W. dip above the horizon, when their amplitude is observed to compare with their true computed amplitude to find the variation. Or the lower limb of the sun ought to be about 17' + the dip above the horizon. To compute the true amplitude, add together the secant of the latitude, and the sine of the object's declination, and the sum rejecting ten from the index will be the sum of the true amplitude, east when the object is rising, and west when setting; north when the declination is north, and south when it is south. Then if the true and observed amplitude, be both north or both south, their difference, otherwise their sum, is the variation; westerly when the true amplitude is to the left of the observed, and easterly when the true altitude is to the right of the observed. Let P, fig. 16, plate II., be the pole, A the east or west points of the horizon, C B or C′ B, the declination of the object at rising or setting, then A C or AC' is the amplitude, BC or B'C the declination, and BAC or B'A C' the colatitude; and rad. sin. BC= sin. A C sin. BAC, r sin. BC r. sin. BC whence sin. AC= sin. BAC sect. lat. sin. declin rad. = cos. lat. Erample. If on Oct. 11th, 1828, in lat. 50° 46′ N., long. 17° W., the sun rise E. 20° S. by compass at 6h. 32m. A. M., required the variation? h. m. 32 Time Oct. 11th 18 Long. Greenw. time 19 40 Variation To compute the true azimuth of any celestial object from its altitude, polar distance, and the latitude of the place of observation. Lat. 50 46 sect. 10-198953 E. 11 11 S. E. 20 0 S. sin. 9.287923 8 49, W. the true bearing being to the left of the observed. rithms will be the sine of half the azimuth, to be reckoned from the north in south latitud, and from the south in north latitude, eastward when the latitude is increasing, and westward when it is decreasing. Then, when the true and observed azimuths are both east or both west, their difference is the variation, otherwise their sum is the variation, westward when the true is to the left, and east ward when it is to the right of the observed. Add together the altitude, latitude, and polar distance, and take the difference between half the sum and the polar distance. Then add together the secant of the altitude, the secant of the latitude, rejecting ten from the index of each, the cosine of the half sum, and the cosine of the remainder, and half the sum of these four logaFor, adopting the notation employed in investigating the rule for computing the meridian distance of a celestial object, we have (fig. 10, plate II.) a+l+p. COS COS. ·p sect. sect. a. 30° 14′ N., long. by account 31° W. at 4h. 30m. P. M. the W. per compass, height of the eye fifteen feet, required From the observed altitudes and the distance of the sun and moon, and the compass bearing of either object, to find the latitude, longitude, and variation of the compass. With the observed distance enter the Nautical Almanac for the time of the month in which the observation is made, and take by inspection the day and hour of Greenwich time, corresponding most nearly with that distance. To that time take the moon's semi-diameter and horizontal parallax, and, clearing the distance, find the Greenwich time from it. If this time differ much from that before taken out by inspection, take out the moon's semi-diameter and parallax again, and, again clearing the distance, find from it the true Greenwich time. For this time take the polar distance of both objects, and proceed with the computation thus : Let M (fig. 17, plate II) be the true place of the moon; S that of the sun; MP, S P, their polar distances; and M Z, SZ, their true zenith distances; and MS their true distance. Then in the triangles P M S, Z MS, all the sides in each 10 W. 37 W. 16 33 W. or nearly 1 W. are given to find the angles PMS, Z MS, the difference of those angles is the a Z MP; and in the triangle Z M P are given M P, and the included angle Z M P, to find ZB the co-latitude. Hence the time at the place observation, and the true azimuth of eithe object, may be found; and the time compan with the Greenwich time, previously found fro the distance, will give the longitude; and the true azimuth compared with the observed o will give the variation of the compass. Example.-On September 2d, in the morning the altitude of was 15° 45′ + of ) 5840 distance of their nearest limbs 77° 0′ 40′, lat by account 48° N., height of the eye twelve feet, bearing of the sun S. E. E., required the lats tude, longitude, and variation of the compass! By inspection in the Nautical Almanac it readily seen that the Greenwich time of this servation must have been about 19h. of Septem ber 1st. To this time the moon's semidiameter is 15'0", and hor. par. 55′ 2′′ h. m. S. 15 45 15 12 Semid. 15 53 |