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sions and declinations of the fixed stars for the motion of light, and the effect of the attraction of the sun and moon on the position of the earth's axis, which corrections are not given in Table 23. But the true apparent right ascensions and declinations of 24 of the principal fixed stars, for every tenth day of the month, are given in a table added to the Nautical Almanac ; and in deducing, by this rule, the error of a chronometer on land by altitudes taken from an artificial horizon, it will be proper to take the right ascensions and declinations from that table in the Nautical Almanac ; but observations taken at sea admit of no such precision in their results as to render these minute corrections an object worth regard.

EXAMPLE.

If on the 20th of February, 1863, at 9h 24m 4s P. M. the altitude of the star Menkar be 13° 50 decreasing, latitude 52° 30' N, longitude 36° W, height of the eye 15 feet, required the apparent and mean time?

9h 27m 4s P.M. time by watch. 2 24 0 W longitude in time. 11 51 4 P. M. Green, time.

*'s R. A. 1820, 2h 52m 53s + 3.12s Cor. for 43 years

2 14 *'s R. A. for 1863 2 55 7

O's R.A. Feb. 20, 1823,

at noon Green. time 22h 12m 368 + 3m 50s *'s dec. 1820.. 3° 22' 40"N+14:59s Red. to 1863, (Tab. 20) 1 13

Cor. for 43 years

10 27 22 13 49

*'s dec. 1863.. 3 33 7 N Cor. for Green. Time.. 1 54

90 0 0 True right ascension.. 22 15 43

Polar distance 86 26 53

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EXAMPLES FOR EXERCISE.

Required the apparent and mean times in the following examples ?

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i loct. 8, 1866 20 26 N 22 w 10 29 10 P.M. Aldebaran ig 60
2 June 2, 1851 27 42 S 15 W 7 13 28 P.M.Sirius 17 25
3 May 28, 1877 59 15 N 174 E 152 16 A.M.Markab 21 25 +
4 Aug. 1, 1345 30 28 S12 E 9 59 41 P.M.Spica 11 15
5 Mar. 8, 1842 12 30 N 62 E 11 24 7P.M. Procyon 42 58
6 July 15, 1873 8 50 S 7 W10 3 OP.M.Arcturns 29 48
7 Dec. 5, 186853 26 s 30 E 250 56 A.M. Spica 11 26
8 Sep. 12, 1823 32 14 N 130 E 9 48 4 P.M. 2

10 17 9 Sep. 16, 1823 19 30 S 61 E 450 15 P.M.

29 3 10 Sep. 29, 1823 38 20 N 30 W 230 26 A.M. J

22 45

+1+111+1++

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To compute the equation of equal altitudes. The equation of equal altitudes is a correction for the change of the sun's declination, to be applied to the middle time between the instants, by a chronometer, at which on a given day the sun has equal altitudes, to find the time by the chronometer when he is upon the meridian.

To compute this correction, add the cosine of half the interval, in degrees, between the times of observation, to the cotangent of the latitude, and the sum, rejecting 10 from the index, will be the tangent of arc first, the difference between which and the polar distance, will be arc second.

Then add together the cotangent of the same half interval, the cosec. of arc first, the cosec. of the polar distance, the sine of arc second, and the logarithm of one thirtieth of the seconds in the sun's change of declination in the interval between the observations, and the sum, rejecting the tens from the index, will be the logarithm of the seconds of time in the equation of equal altitudes.

Note. This rule is an approximate one, but the result will always be correct to within a fraction of a second. In performing the latter part of the calculation, the first four decimals in the logarithms are all that are necessary; and no great exactness is requisite in either the latitude or polar distance, the latter of which, without producing any error worth notice, may be deduced from the declination for the nearest noon.

EXAMPLE. On July 25, 1823, in latitude 54° 20° N, at 8h 59m 4s A. M., and 3h Om 40s P. M., the sun had equal altitudes, required the equation of equal altitudes? 20h59m 4s

O's declination at noon, Greenwich time, 27 0 40

20° 1' – 12' 35". The proportional part of Interval .... 6 1 36

12' 35" corresponding to 6h Im is 194s, zóth Half interval 0 48=45° 12' of which is 6.5"; and the polar distance is

69° 59'.

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In each of the following examples the equation of equal altitudes is required

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To find the error of a chronometer by equal altitudes of the sun.

Two or three hours before noon, let several altitudes of the sun's upper or lower limb be taken, noting the time of each observation by the chronometer, and again in the afternoon let the successive times by the chronometer be noted at which he arrives at the same altitudes. Take the mean of the times at each set of observations, adding 24 hours to the mean of those in the afternoon, and half the sum of these means will be the middle time between the observations, and the difference of them will be the elapsed interval of time. With this interval, the latitude, polar distance, and change of declination, compute the equation of equal altitudes, as in the last problem, and add it to the middle time, if the polar distance is increasing, but subtract it if decreasing, and the result will be the time shown by the chronometer at apparent noon, and the difference between that time and 24 hours will be the error of the chronometer for apparent time at the place of observation.

To find its error for mean time, either at the place of observation or at Greenwich, reduce the equation of time to the given meridian, (see the use of Table 22,) and add it to, or subtract it from, 24 hours, as directed in the Nautical Almanac, or Table 22, and the result will to which apply the longitude of the place in time, adding it if west, and subtracting it if east, and the mean time at Greenwich when it is apparent noon at the given place will be obtained. Hence the difference between the middle time, corrected as above for the equation of equal altitudes, and the mean time at the place when it is apparent noon, will be the error of the chronometer for mean time at the place of observation; and the difference between the corrected middle time and the corresponding mean time at Greenwich will be the error of the chronometer for Greenwich mean time.

EXAMPLE.

On July 25, 1823, in latitude 54° 20' N, longitude 20° 30 E, equal altitudes of were observed, as under; required the error of the chronometer for apparent and mean time at the place of observation, and also for mean time at Greenwich ?

Add. 8h 58m 30s A.M. 31°30'Q. 3h Om 6s P. M. Equation of time Oh 6m78 + 1s 58 47

35 0 23 Cor. for long .. 0 0 0
59 4

40 040 Reduced equa.. 0 6 7
59 21
45 0 56

Time of ap.noon 24 0 0
59 38

50
1 15

Mean time at do, 24 6 7
Mean
8 59 4

3 040 Long. in time E 1 22 0 12

24

Mean T. at Gr. 22 44 7 Mean 20 59 4

27 40 times 27 0 40 Sum .. 47 59 44 half sum 23h 59m 52s mid, time. Diff ..

1 36 half diff. 3 0 48 45° 12' half interval.

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The equation of equal altitudes (see the last problem) is about 10s; and as the polar distance is increasing, it must be added to the middle time. Hence 23h 59m 52s + 10s = 24h Om 2s, the time by the chronometer at apparent noon.

Time by chronometer at apparent noon ....
Time of apparent noon
Mean time at apparent noon
Corresponding Greenwich mean time
Chronometer fast for apparent time at place of observ..
Chronometer slow for mean time at place of observ....
Chronometer fast for mean time at Greenwich.

24h Om 2s
24 0 0
24 6 7
22 44 7

2 6 5 1 15 55

EXAMPLES FOR EXERCISE.

In the following examples the error of the chronometer for mean time at Greenwich is required, the sun having in each case equal altitudes at the times marked A. M. and P. M.?

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h m 1 Sep. 1, 1824 8 7 29 A.M.3 54 4P.M.46 50 N 32 ze

2 9 40 fast. 2 Mar. 5, 1825 7 45 21 4 14 49 38 34 N 59 30 W 4 9 55 slow. 3 Apr. 17, 1822 9 15 43 2 45 0 32 18 N55 50 W 3 42 39 slow. 4 Nov. 27, 1836 10 7 22 1 52 26 23 4 N 16 4W0 52 17 slow. 5 Jan. 12, 1849 8 28 26 3 31 12 8 56 N51 14 E 3 15 59 fast. 6 Apr. 25, 1870 9 30 35 2 29 16 16 OS 13 OE 0 54 11 fast. 7 Sep. 4, 1842 8 22. 18 3 37 39 124 30 S 47 OW 3 7 6 slow. 8 Sep. 9, 1832 9 7 42 2 528 61 26 N22 OE 1 31 16 fast.

To find the rate of a chronometer. The rate of a chronometer, or its daily gain or loss, is determined by comparing its errors for mean time, as found by observation at a given place on different days, either by the altitude of the sun or a star, or by equal altitudes of the sun, as shown in the preceding problems. Thus if by observation a chronometer is found 20s fast, and at the end of ten days 30s slow for mean time at the same place, it has evidently lost 50s in ten days; whence its daily rate is 5s losing. If on a given day a chronometer be 12s fast, and at the end of thirteen days 57s fast for mean time at any place, it must have gained 45s in thirteen days, or its rate is about 3.5s per day losing. Hence the method of finding the rate of a chronometer is evident.

When the longitude of a place at which observations are taken for the error of a chronometer is accurately known, it will be found convenient in all cases to find its error for Greenwich time, which may be done by finding (Problem 4, page 219) the time at Greenwich corresponding to the time at the given place as computed from observation, and comparing that time with the time shown by the chronometer. When the errors are all referred to Greenwich time, the rate can easily be found from observations taken at different places ; and it is often desirable to do this; for a seaman cannot always stop long enough at one place to obtain a rate that is entitled to confidence, as the unavoidable errors of any observation render it advisable that a rate should not be deduced from observations separated from each other by a very short interval of time. If more than two sets of observations can be obtained, it will be seen from the results whether the rate of the chronometer is uniform or not; and for this purpose the mariner should lose no opportunity to multiply observations, that any change in the rate may be detected and allowed for.

The student, if acquainted with the preceding problems, will, from what has been said, experience no difficulty in finding the rate in each of the following examples, or indeed in any case that can arise

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