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To find the apparent time at Greenwich. True distance

540 0' 37"
Distance at noon, Sept. 27, 54 20 29 first difference 0°19' 52" P. log. 9571
Distance at

. . 3h 52 34 58 second difference 1 45 31 P. log 2320
September 27.. 0 33 54 P. log 7251

24 0 0
Greenwich time, September 26, 24 33 54 ; for the astro-
nomical time at the place of observation being on the 26th, the time at Greenwich
must be estimated from the noon of the same day.

To find the apparent time at the place of observation.
O's R. A. September 27, at noon, Greenwich time 12h 13m 2s + 3m 37s
Correction for Oh 34m past noon.

0 0 4 O's reduced R. A......

12 13

6

+

*'s R. A. Jan. 1, 1820, 9h 58m 47s + 3.21s Declination.... 12° 50'36" N - 17.23s 3.21s x 3 years = ..

12
17.23x38 years

1 4 *'s reduced R. A. 9 58 59

*'s reduced dec. 12 49 32 N

90 0 0 *'s polar dist. 102 49 32

*'s true altitude.. 10°31' 37" Latitude

28 37 0 sec 056583 *'s polar distance 102 49 32 cosec .010973 Sum..... 2)141 58 9 Half sum

70 59 4 cos 9.512985 Remainder

60 27 27 sin 9.939507

2)19.520048 Half hour angle 35° 8' sin 9.760024

8 *'s meridian dist. 4h41m 4s *'s R. A..

9 58 59 R. A. meridian.... 5 17 55 O's R. A.... 12 13 6

17 4 49 apparent time at the place of obser on.

24 33 54 apparent time at Greenwich. Difference

7 295, longitude in time W, 112° 164 W.

1

EXAMPLE III.

On September 16, 1823, at noon, we were in latitude 28° 27' N, and longitude by account 40° W; at 4h 1m 2s P. M. per watch, the altitude of Q was 30° 14' -, and at “h 40m 2s the distance of )'s nearest limb from Antares was 66° 30' 12", the ship having run from noon 'S WW 5 miles an hour, height of the eye 12 feet, required the longitude when the distance was measured ?

The distance run from noon till 4h P.M. is 20 miles, and till 7h 40m

difference of latitude made from noon till 4h P. M. is 13, and till 7h 40m, 24 miles, consequently the latitude, when the O's altitude was observed for the time, was 28° 14' N, and when the distance was observed 28° 3' N.

Again, from 4h P.M. till 7h 40m the ship had run SW1W 18 miles, whence the departure which she had made in that time is 14 miles nearly, with which, and the middle latitude, nearly 28°, the difference of longitude which she has made is 1m 4s W.

To find the error of the watch from O's altitude.
Time per watch ..... 4h lm 2s O's declination at noon 2°51'58"N - 23' 11"
Long. in time by acct. 2 400 Cor. for Green. time... 0 6 27
Green. time by acct... 6 41 O's dec. at given time 2 45 31

90
O's polar distance 87 14 29

2

..

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Time per watch when distance was observed .....

7h 40m2 s Watch fast for time at the meridian on which the ship was at 4h P. M. 0 14 14 Apparent time at that meridian

7 25 48 Difference of longitude in time made since 4h P. M....

0 14 Apparent time at the meridian on which the distance was taken 7 24 44 Longitude by account in time

2 40 0 Greenwich time by account when the distance was observed

10 The right ascension of the sun, and the right ascension, declination, &c. of the moon, and the star being reduced to this time, we have O's R. A. 11h 35m Os, )'s R. A. 21h 2m 52s, )'s P. D. 105° 15', *'s R. A. 16h 18m 33s, *'s P. D. 116° 2', )'s horizontal semidiameter 14' 52", and parallax corrected by Table 14, 54' 30",

4

Apparent time.... 7h 24m 44s

18h 59m 44s
O's R. A.
11 35 0

16 18 33 *'s R. A. R. A. mer...

18 59 44 *'s mer, distance 2 41 11
D's R. A.

21 2 52
D's mer. distance.. 2 3 8 30° 47'

40° 18'

To compute the moon's altitude.

Latitude.... 28° 3' cot •273412 sin 9•672321
D's mer, dist. 30 47 cos 9.934048
Arc 1

58 12 tan 10.207460 sec .278226
) 's pol. dist. 105 15
Arc 2....... 47 3

cos 9.833377 )'s hor. semid... 14' 52" D's true altitude. 37° 27' 0" sin 9.783924 Aug.......

0 0 9 Correction nearly

42 0

)'s true semidiam. 0 15 1 Apparent altitude nearly . 36 45 0

) 's N. L. from * 66 30 12 True correction..

42 24

App. central dist. 66 45 13 Apparent altitude

36 44 36

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Difference of apparent altitudes 13° 14' 0"
Auxiliary arc...

60 17 50 Apparent distance..

66 45 13 Sum of auxiliary arc and preceding one 73 31 50 suvers 83458 Difference of ditto

47 3 50 suvers 81147 Sum of arc first and following one 127 3 3 vers 02512 Difference of ditto ..

6 27 23 vers

06330 Difference of true altitudes

13 58 35 vers

29564

03011 Sum of parts for "

146 Vers true distance

03157 66°37' 011

03119 8

38 True distance .... 66 37 8

Parts for

46 36 12 12 40

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true distance,

To find the apparent time at Greenwich.

7488
•2987
4501

True distance...... 66°37' 8".
Distance at 9h 66 5 2 first difference 0°32' 6" prop. log.
Following distance .. 67 35 31 second difference .. 1 30 29 prop. log

1h 3m5ls prop. log
Time of first distance

9 0 0 Apparent time at Greenwich

10 3 51 Apparent time at ship ....

7 24 44 Longitude in time

2 39 7 39° 463! W

EXAMPLES FOR EXERCISE.

In the following examples the time at the place of observation is deduced from O's altitude observed with the distance, the longitude being required ?

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2

32 35 2 32 462 78 5

25 45
2 62 37

13 26+
J 61 45
2 22 39
2 43 51

15 24+
2 37 45
0 27 59-
J 70 41

37 16+
2 50 35

23 36+

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In the following examples the time at the place of observation is deduced from *'s altitude observed with the distance, the longitude being required

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In the following examples the true apparent time at the place of observation is given, and the altitudes for clearing the distance are to be computed, the longitude being required ?

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INVESTIGATION OF THE TRIGONOMETRICAL FORMULÆ
WHICH FORM THE PRECEDING RULES FOR THE
PRACTICAL SOLUTION OF THE MOST USEFUL PRO-
BLEMS IN NAUTICAL ASTRONOMY.

Method of finding the latitude from two altitudes of the sun, and the time

elapsed between the observations. (See p. 230.)

LET Z or Z' be the zenith, ID CH the horizon, P the pole, A and B the planes of the sun at the two times of observation, A D and B C the true altitudes, A Z, BZ, or A Z, BZ' the zenith distances, A Pand B P the polar distances, which, in the practice of this problem, may be considered as equal. Join A B, and bisect it by the

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