EXAMPLE. If the apparent altitude of the moon's lower limb be 40° 26' in latitude 58°, her semidiameter, from the Nautical Almanac, being 16' 2", and horizontal parallax 58' 47", the index error of the instrument 5' 2" +, and height of the eye 14 feet, required the true altitude ? D's semidiameter 16' 2'' Horizontal parallax 58' 47" 40° 26' 0'' observed altitude 2. TABLE 24. + 5 2 index error. Min. D's Parallax 58' 40 31 2 3 41 dip. 4 40 27 21 30%.. 22 Sec parallax { + 16 13 semidiameter. 6. 4 40 43 34 Cor ) 's app alt 43 17 43 17 correction of altitude. 41 26 51 true altitude. EXAMPLES FOR EXERCISE. In each of the following examples, the moon's true. altitude is required ? Height Answer. Lat. Observed Alt. Index Error. Hor.Paral. Semidiam. of the Eye. True Alt. Feet. Å8 2 53 18 20 + 5 58 14 14 54 9 34 20 2 49 2 48 4 26 60 27 16 29 19 49 49 28 67 J 57 14 23 3 11 54 30 14 52 20 57 20 46 32 263 42 55 29 15 8 12 64 22 3 10 J20 4 57 2 0 54 14 14 47 16 20 32 34 39 2 6 3 28 + 1 48 23 7 7 55 + 4 12 PROBLEM IX. To find the polar distance of a celestial object. When the declination of the object and the latitude of the place are of contrary names, add the declination to 90°; but when they are of the same name, subtract the declination from 90°, and the sum or remainder will be the polar distance. EXAMPLES. Lat. Declination. Polar Dist. To deduce the latitude from the observed meridian altitude of a known celestial object. S" If, in the annexed figure, A C be the intersection of the plane of the meridian and that of the rational horizon, Z and F D its intersection with the plane S" of the equator; and if Z be the zenith, P the north and E the south pole, and S the true place of a celestial object on the meridian; then SA is its meridian А altitude, S Z its meridian zenith distance, and SF its' north declination; and if s' be the place of the object, S' A is its alti- E D tude, S Z its zenith distance, and S F its south declination ; if S" be the place of the object, s”C is its altitude, S" Z its zenith distance, and S'' E its north declination. Now ZS + SF, or S2 - SF, or SF S" Z = ZF the latitude. Hence if the object be the sun, correct his altitude, as in Problem 7, page 222, and reduce his declination to the given day, and the given meridian, as shown in the use of Tables 19, 20, 21, and 22. If the object be the moon, find the Greenwich time of her passing the meridian, as in Problem 5, page 220, or in the use of Table 17, and to that time (see the use of Table 30,) reduce her declination, semidiameter, &c. and compute her true altitude, as in Problem 8, page 223. If the object be a planet, its declination may be taken from the Nautical Almanac, where the declination is given for every sixth day in the month ; but more conveniently from Shumacker's Ephemeris, where the declinations, &c. of Venus, Mars, Jupiter, and Saturn, are given for the noon of every day, Greenwich time; and to its altitude, computed as if it were a fixed star, the parallax from Table 26 may be added to obtain its true altitude. If the object be a fixed star, let its declination be reduced to the given instant from Table 23, and its true altitude computed as in Problem 6, page 221. Then, whatever be the object, the complement of its true 'altitude will be its true zenith distance; which is to be called north or south, according as the object is south or north of the zenith ; and when the zenith distance and declination are both north, or both south, their sum is the latitude; but when one is north, and the other south, their difference is the latitude, and of the same denomination with the greater. Note. It is perhaps difficult, under any circumstances, to obtain an altitude at sea, which can be depended on, to much nearer than a minute; it will therefore, in general, be sufficiently exact in sea prac + tice, to take the semidiameter and other corrections to the nearest minute ; and in finding the latitude at sea, the small corrections for the parallax of the sun and the planets may be safely disregarded EXAMPLE I. If the meridian altitude of @ be 56° 10 40" S, by a back observation, on May 4th, 1845, in longitude 31° W, height of the eye 20 feet, required the latitude ? 31°W 56° 10' 40" alt. QS. 4 + 4 24 dip. 60) 124 56 15 4 38 refraction. 2h 4 longitude in time. 56 14 26 l's dec. for noon, Greenwich time, + 15 52 semidiameter. May 4, 1825 15°57' 37" + 17' 15" Reduction to 1845, Table 20.. + 2 33 56 30 18 5 parallax. 16 00 10 Correction for longitude, Table 30 1 29 56 30 23 true altitude. 90 True declination ... 16 1 39 N 33 29 37 zenith dist. N. 16 1 39, declination N. 49 31 16 latitude N. EXAMPLE II. On September 19th, 1823, in longitude 51° E, the meridian altitude of 2 was 36° 52' N, height of the eye 30 feet, required the latitude ? Per Nautical Almanac, (see Table 31,) D's mer pass at Greenwich 11h 37m + 44 Correction for longitude 50° E and 44m, (Table 17,) 6 11 31 Longitude in time E 3 24 Time of obseryation at Greenwich 7 0 35 11 N 2 46 44 prop. log. 332 Double Greenwich time 16h 14m Table 30 1698 Change of ) 's declination in 8h 7m 1° 52' 47" prop. log. 2030 Declination at noon ....... 2 11 34 S Declination at time of observation...... 0 18 47 S The horizontal parallax and semidiameter reduced in the same way, are 15' 13" and 55' 49''. Observed altitude 2 36 52 0"N Semidiameter 15' 13"] + 15 22 Aug. ........ 9 37 7 22 Dip. 5 24 37 1 58 37 45 16 90 Zenith distance... 52 14 44 S Declination 0 18 47 S 8 + 43 18 EXAMPLE III. If the meridian altitude of Rigel be 85° 6' N, on November 25th, 1848, height of the eye 20 feet, required the latitude ? *'s dec. Jan. 1, 1820, (Table 23,).... 8°25' 2"S Observed alt. 85° 6' 0"N Annual var. 4":74 X 28:9 years 2 17 Dip 4 24 Declination at time of observation.... 8 22 45 S 85 1 36 Refraction .. 5 85 1 31 90 0 0 Zenith dist... 4 58 29 S Declination.. 8 22 45 S 13 21 14 S Latitude .... EXAMPLES FOR EXERCISE. In each of the following eramples the latitude is required ? To find the latitude from the observed altitude of a known celestial object when on the meridian below the pole. In the last figure F P and Z C are quadrants, if therefore the common part ZP be omitted from each, PC, the elevation of the pole above the horizon, will be left equal to F Z the latitude. Hence if $'" be an object on the meridian below the pole, S'' C its altitude, added to S'" P the complement of its declination, will be equal to the latitude, and of the same name with the declination. Now the sun is on the meridian below the pole 12 hours after noon, therefore if the longitude in time be added to or subtracted from 12 hours, according as it is west or east, the Greenwich time at which the sun will be below the pole will be obtained. If to the time of the moon's passing the meridian, determined by Problem 5, page 220, 12 hours, and half the daily difference of her meridian passage be added, the sum will be the time at which she will be below the pole ; and if 12 hours be added to the time at which a planet is on the meridian, the sum will be the time at which it is below the pole, exactly enough to compute its declination. Let then the declination of the object be taken out for the instant of Greenwich time at which it passes the meridian below the pole, and to the complement of its declination add its true altitude, and the sum will be the latitude, of the same name with the declination, EXAMPLE. If the altitude of 0 on the meridian below the pole be 6° 26', on July 4, 1822, longitude 21° W, height of the eye 18 feet, required the latitude ? h m 12 0 O's declination at noon .. 22° 56' 5"N - 5' 17" Long. in time W + 1 24 Cor. for Green. time, Tab.30 - 2 56 Greenwich time 13 24 Reduced declination...... 22 53 9N 90 Co-declination 67 6 51 Observed altitude o 6° 26' 0" Dip 4 11 6 21 49 Correction for parallax and refraction.... 7 54 6 13 55 Semidiameter 15 46 True altitude 6 29 41 Co-declination... 67 6 51 73 36 32 N To find the latitude from the altitude of the Pole Star at any given time. The distance of the pole star from the pole is so small, that the |