Page images
PDF
EPUB

Moon's right ascension reduced to Greenwich time =

Moon's horary distance, west of meridian

Right ascension of the meridian =

[ocr errors]
[ocr errors]
[ocr errors][merged small][merged small]

Mean sun's corrected right ascension

[ocr errors][merged small][merged small][merged small][merged small]

Correct mean time at the given meridian

Note. In the same manner may the mean time at ship, or place, be deduced from the horary distance of a planet.

PROBLEM XXI.

Given the Mean Time under a known Meridian; to find the Sun's Horary Distance from the Meridian :-or, to convert Mean Time into Apparent Time.

RULE.

This problem is merely the converse of Problem XIX., page 369; -hence, let the equation of time, page II. of the month in the Nautical Almanac, be reduced to the mean time at Greenwich:-then, this reduced equation being applied to the given mean time according to its sign in that page; the result will be the sun's horary distance from the meridian; or, in other words, the apparent time at the given meridian.

Example.

February 1st, 1836, at 3:10:18: mean time, in longitude 50:40?· west; required the sun's horary distance from the meridian?

[merged small][ocr errors][merged small][merged small][merged small][merged small]

time, page II. of the Naut. Alm., being reduced to the time at Greenwich; the result will be, to the nearest second 343 subtractive. Given mean time =

[ocr errors]

Sun's horary distance from the meridian = rent time from noon.

3:10:18:

3 635; or the appa

Note. The apparent time only relates to the sun; it does not apply to either the moon, the fixed stars, or the planets.

See Explanatory Article 36, page 313, relative to applying the equation to mean time, and conversely.

PROBLEM XXII.

Given the Mean Time under a known Meridian; to find the Meridional Horary Distance of the Moon, a Fixed Star, or a Planet.

RULE.

To the given mean time at ship, or place, apply the longitude in time, agreeably to the directions contained in Problem III., page 342: the result will be the corresponding mean time at Greenwich; to which time, let the mean sun's right ascension (viz., the "Sidereal Time," in page II. of the month in the Ephemeris), be reduced by Problem V., page 344.

To the mean sun's reduced right ascension, add the mean time at ship, or place, and the sum, abating 24 hours, if necessary, will be the right ascension of the meridian.—(Problem VI., page 345.)—Now, the difference between this and the corrected right ascension of the given celestial object, will be the required horary distance from the meridian which will be east when the right ascension of the given object is greater than the right ascension of the meridian; otherwise it will be west.

It must be remembered that the right ascension of the moon is to be reduced by Problem XVI., and that of a planet by Problem XVII. -The right ascension of a star is to be taken from the Nautical Almanac (between pages 368 and 407), and reduced to the given day.

Example 1.

May 1st, 1836, at 10:20 28 correct mean time, in longitude 70:36: east; required the horary distance of Antares from the meridian?

Mean time at ship. 10:20:28: || Mean sun's R. A. at

[merged small][merged small][merged small][merged small][merged small][ocr errors][merged small][ocr errors]

noon, May 1st. 2:37:46:20 Equation to 5:38"4:

in Table XLVI. +0.55.53

[ocr errors]

Mean sun's cor. R. A. 2:38" 41:73

Right ascension of the meridian, or sidereal time

Right ascension of Antares

Star's horary distance from the meridian.

[ocr errors][ocr errors][merged small][merged small][ocr errors][ocr errors][merged small][merged small]

which is east, because the star's right ascension is greater than the

right ascension of the meridian.

Example 2.

May 3rd, 1836, at 17:44:31: correct mean time, in longitude 76:42 east; required the moon's horary distance from the meridian?

Mean time at ship = 17:44:31 || Mean sun's R. A. at

[merged small][merged small][merged small][merged small][ocr errors][merged small][merged small][merged small][merged small][merged small][merged small][ocr errors]

Right ascension of the meridian, or sidereal time =
Moon's corrected right ascension=

Moon's horary distance from the meridian =

20:32 14:79 -17. 19. 41.92

. 3:12:32:87;

which is west, because the moon's right ascension is less than the right ascension of the meridian.

The above is simply the converse of Problem XX., page 370.

Remark. The reason of the horary distance being called east, in the first example, and west in the second, will appear obvious by reflecting that as the right ascension of the meridian signifies that point of the Equinoctial which comes to the meridian of a place at any given time; and, as that point expresses the right ascension of the heavens, reckoned in sidereal time from the first point of Aries; therefore, any star or other celestial object, whose right ascension is greater than the right ascension of the said point of the Equinoctial at a given time, is to the eastward of the meridian which it represents; and vice versa, any star, or other celestial object, whose right ascension is less than the right ascension of the said point, is to the westward of the same meridian. Let the young navigator attend to this remark, and he will never be at a loss to know the true position of a heavenly body at any time, with respect to the meridian of a given place.

PROBLEM XXIII.

Given the Observed Altitude of the Lower, or Upper Limb of the Sun; to find the true Altitude of its Centre.

RULE.

To the observed altitude of the sun's lower limb * (corrected for

* See Articles 60 and 61, page 326.

index error, if any), add the difference between its semidiameter* and the dip of the horizon; † the sum will be the apparent altitude of the sun's centre: or, from the corrected observed altitude of the sun's upper limb subtract the sum of the semidiameter and the dip of the horizon; and the remainder will be the apparent central altitude,

Now, from the apparent altitude of the sun's centre, thus found, subtract the difference between the refraction ‡ corresponding thereto, and the parallax in altitude,§ and the remainder will be the true altitude of the sun's centre.

Note.-The difference between the refraction and parallax constitutes the correction of the sun's apparent altitude in the lunar observations.

Example 1.

Let the observed altitude of the sun's lower limb, by a fore observation, be 16:29, the height of the eye above the level of the sea 24 feet, and the sun's semidiameter 16:18"; required the sun's true central altitude?

Observed altitude of the sun's lower limb =
Sun's semidiameter =

[ocr errors]

16:29: 0

16:18"

Diff.

Diff.+ 11.36

16:40:36"

[blocks in formation]

3. 0

True altitude of the sun's centre =

Example 2.

16:37:36

Let the observed altitude of the sun's upper limb, by a fore observation, be 18:37, the height of the eye above the surface of the water 30 feet, and the sun's semidiameter 15:46"; required the true central altitude?

Observed altitude of the sun's upper limb =

Sun's semidiameter =

Dip of the horizon for 30 feet =

18:37: 0:

[ocr errors]

15:46

5.15.J

Sum-21. 1

[merged small][ocr errors][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small]

* Page II. of the month in the Nautical Almanac. + Table II.

§ Table VII,

+ Table VIII.

Remark. As the back observations (given in the first impression of this work) are more curious than useful; and, on the whole, subject to so many inconveniences in practice as to be very rarely, if ever, resorted to for the purpose of finding the latitude of a ship; they are, consequently, omitted in the present edition, as points not deserving the mariner's attention.

The reader is now requested to observe that throughout the rest of this work the altitudes of the heavenly bodies will be adapted to the fore observations, the same as in the above examples; because these observations are the most natural, the most simple, and the most convenient to the practical navigator.

PROBLEM XXIV.

Given the Observed Altitude of the Upper, or Lower Limb of the Moon ; to find her true Central Altitude.

RULE.

To the given mean time at ship, or place, apply the longitude, in time, agreeably to the directions contained in Problem III., page 342; the result will be the mean time at Greenwich; to which let the moon's semidiameter and horizontal parallax be reduced by Problem XV., page 361 (or by Table XVI., as explained between pages 27 and 29), and let the reduced semidiameter be increased by the correction contained in Table IV., answering to it and the observed altitude :-then,

To the observed altitude of the moon's lower limb* (corrected for index error, if any), add the difference between the true semidiameter and the dip of the horizon; or, from the observed altitude of the upper limb subtract the sum of the semidiameter and dip, and the apparent central altitude of the moon will be obtained; to which let the correction (Table XVIII.) answering to the moon's reduced horizontal parallax and apparent central altitude be added, and the sum will be the altitude of the moon's centre.

Example 1.

In a certain latitude, March 10th, 1836, at 13:40 20:, mean time, the observed altitude of the moon's lower limb was 20:10:40", and the height of the eye above the level of the sea 24 feet; required the true altitude of the moon's centre; the longitude being 35:40: west?

* See Article 62, page 327.

« PreviousContinue »