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EXAMPLE 1.

1. Required the moon's declination March 17th 1822, at 7h.22′ in longitude 57° W.

The moon's declination at noon (Naut. Alm.) at Greenwich, being 26°.12′ S. and at midnight 24°.58' S.

Time at the given place
Longitude 57° W. in time

Time at Greenwich

>'s declination at noon

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7h.22'

= 3.48′ W.

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D's declination at midnight24°.58′ S.

Decrease in 12 hours

- = 1°.14'

Hence 26°.12'-1°.9′-25°.3′ S. the moon's declination at the time and place required.

EXAMPLE II.

Required the moon's semidiameter and horizontal parallax on the 26th of January 1822, in longitude 11°.45′ West, at 15h.45' apparent time.

The semidiameter, at Greenwich, at midnight, (Naut. Alm.) being 16′.1" and at noon on the 27th, 16'.3", also the horizontal parallax at the same time 58'.32", and 58'.41".

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D's semi-dia. at midnight, 26th,= 15'.57"; hor. paral.

D's semi-dia. at noon,

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27th, 15.59"; hor. paral.

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12h: 9":: 4.32′; 3′4

15.57". hor par. at midnight

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▷'s ditto in long. 11°.45′ W.=15′.57".7 hor.par.in long.11°.45′W.=58′.35′′.4

Hence the 's semi-diameter at 15h.45' in long. 11°.45' W. is 15'.57.7, and the horizontal parallax is 58′.35".4.

given in the VIth page of the Nautical Almanac: and her latitude, longitude, right ascension, declination, semidiameter, and horizontal parallax, are given for noon and midnight at Greenwich in pages Vth, VIth, and VIIth, for each month. * If the time at Greenwich had exceeded 12 hours, the moon's declination must have been taken out for midnight and the noon of the next day; and the variation applied to the midnight declination.

PRACTICAL EXAMPLES.

1. Required the moon's declination on the 11th of January, 1822, at 17b.47′ in longitude 162° West?

The moon's declination at noon at Greenwich (Naut. Alm.) on the 12th of January, being 10.19′ N. and at midnight 1°.35' S.

Answer. The time at Greenwich is 4.35' on the 12th of January, and the moon's declination is 0°.13′ North.

2. Required the moon's semidiameter and horizontal parallax on the 19th of May, 1822, in longitude 38°.40′ E. at 11h.15' apparent time.

The moon's semidiameter at Greenwich (Naut. Alm.) at noon and midnight being 16′.38" and 16'.41′′; and the horizontal parallax at the same time 61'.3" and 61'.13".

Answer. The time at Greenwich is 8h. 40'.20", D's semidiameter 16.40" and horizontal parallax=61'.10".

3. Required the moon's declination on the 13th of May, 1822, at 19h in longitude 67° East.

The moon's declination at Greenwich (Naut. Alm.) at midnight being 15°.5' S. and at noon on the 14th 12°.27' S.

Answer. The time at Greenwich is 14h.32', D's declination 140.32' S.

4. What is the moon's declination on July 19th, 1822, at 4.49′ in longitude 114° East?

The moon's declination at Greenwich (Naut. Alm.) at midnight on July 18th being 21°.12′ N. and on July the 19th at noon 18.51′ N.

Answer. The time at Greenwich is 21h.13' July 18th, and the moon's declination 19°.24' N.

PROBLEM VII.

(D) To find the time of a star's culminating, or coming to the meridian of Greenwich.

RULE. Subtract the right ascension of the sun for the given day from the right ascension of the star, and the remainder will be the time of the star's culminating nearly.-If the sun's right ascension exceed the star's, add 24 hours to the star's before you subtract.

Take the increase of the sun's right ascension in 24 hours, and add it to 24 hours. Then,

This sum is to 24 hours as the star's right ascension diminished by the sun's, is to the time of the star's culminating. NOTE. If the time of culminating be required for any other

T

meridian than that of Greenwich; allow 10 seconds* of time for every 15° of longitude, which subtract from the time at Greenwich for places in west longitude, or add to that time for places in east longitude, and the result will shew the time of culminating at the given meridian.

EXAMPLE I.

At what time will Arcturus come to the meridian of Greenwich on the 1st of December 1822? The right ascension of Arcturus being 14b.7.32".+

*'s right ascension (1822)+24h

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O's right ascension, December 1st, 1822

Time of *'s culminating nearly

38b. 7.32". =16h28.16.

=21b.39.16".

O's right ascension, December 1st, 1822, at noon=16h.28'.16". O's right ascension, December 2d, 1822, at noon=16h.32′.35′′.

Increase of the O's right ascension in 24 hours 0b. 4'.19".

Then 24h. 4.19" 24h::21h. 39'. 16" 21h.35'.23"+ apparent time of Arcturus's culminating, or 9h.35'.23", December 2d, in the morning.

NOTE. Arcturus will be on the meridian of Philadelphia, longitude 75°.13'W,on the same day at 21h.34.33", Philadelphia time; for 15°: 10"::75°. 13′ 50′′ and 21h. 35′.23" - 50"— 21h.34.33". It will also be on the meridian of Pekin, longitude 116°.24' E, on the same day at 21h.36'.40", Pekin time; for 15°: 10":: 116°.24′: 77′′=1′.17′′, and 21.35′.23′′ + 1′.17′′— 21h.36.40". Hence it appears that the same fixed star is on the meridian of every place, in any kingdom of moderate extent, nearly at the same hour reckoned at that place.

PRACTICAL EXAMPLES.

1. At what time will Aldebaran culminate at Greenwich on the 20th November 1822?

The right ascension of Aldebaran being 4.25′.43′′ (Tab. VIII.)

*For the sun's right ascension varies about four minutes of time every day, but the right ascension of a star remains nearly the same during the whole year, and 24h: 4'::1h (=150): 10".

+ See the table annexed to the Nautical Almanac for the year 1822, or Table VIII. of this work.

The same fixed star is on the meridian of any place at nearly the same hour, on the same day of the month, for several years; the variation in 40 years will seldom exceed two minutes of time.

the sun's right ascension at noon (Naut. Alm.) 15h.41'.29" on November 20th, and on the 21st 15h.45'.40".

Answer. 12h.42′.1′′.*

2. At what hour will Regulus culminate at Greenwich, on the 6th of February 1822?

The right ascension of Regulus being 9h.58'.53" (Tab.VIII.); O's right ascension Feb. 6th (Naut. Alm.) being 21.18'.40", and on the 7th 21h.22′.40′′.

Answer. 12h.38'.6".

3. At what time, on the 18th of December 1822, will Sirius, the Dog Star, culminate at Greenwich?

The right ascension of Sirius being 6h.37.18" (Tab. VIII.); O's right ascension 18th December, (Naut. Alm.) at noon, 17.42.55", and on the 19th, 17.47.21".

Answer. 12h.52'.

4. At what time, on the 1st of December 1822, will Castor culminate at Greenwich?

The right ascension of Castor being 7h.23.13" (Tab.VIII.); O's right ascension (Naut. Alm.) December 1st-16h.28'.16", and on the 2d=16h.32.35".

Answer. 14.52′.16′′.

PROBLEM VIN.

(E) To find the time of the moon, or any planet's culminating.

RULE 1. Subtract the sun's right ascension at noon, from the right ascension of the planet, and the remainder is the time of culminating nearly.

If the sun's right ascension, in time, be greater than that of the planet, add 24 hours to the planet's right ascension before you subtract.

2. Find (from the Nautical Almanac) the, daily variation of the sun's right ascension, and the daily variation of the planet's right ascension, and take their difference or sum. When the daily variation of the planet's+ right ascension is greater than that of

* The motion of clocks or watches may be examined, and their errors rectified, by the culminating of the stars. For instance, Aldebaran is on the meridian of Greenwich, November 20th, at 12h.42′.1" apparent time, from which take the equation of time on the same day 14'. 12", as directed in the IId page in the month in the Nautical Almanac, and the remainder 12h.27'.49'' is the true mean time which the clock ought to shew when the star is on the meridian.

+ The right ascensions of the planets are not given in the Nautical Almanac, but they may be calculated from their geocentric latitudes and longitudes, which are inserted in that work, and the obliquity of the ecliptic. The time at which each of the planets passes the meridian of Greenwich is given in the IVth page of each month in the Nautical Almanac, and that of the moon in the VIth.

the sun's, its motion is progressive; and when less, its motion is retrograde.

Then, for the moon, or the progressive motion of a planet. 24 hours diminished by the aforesaid difference: 24h:: the time of culminating nearly the true time of culminating.

For the retrograde motion of a planet.

24 hours increased by the aforesaid sum: 24h:: the time of culminating nearly the true time of culminating.

NOTE. When the planet is stationary, the time of its passage over the meridian will evidently be determined in the same manner as that of a fixed star having the same right ascension with the planet.

EXAMPLE.

Required the time of the moon's culminating at Greenwich on the 13th of August 1822.

>'s right ascension at noon 13th August 1822=94°.59′ =6h.19′.56′′
O's right ascension at noon 13th August 1822

Time of culminating rearly

=9h.30' 4"

20.49′.52′′

=

's right ascen. Aug. 13th=9h30′. 4" D's right ascension Aug. 13th 94°.59' 's right ascen. Aug. 14th = 9'33.50"'s right ascension Aug. 14th=110°.41′ Variation in 24 hours

=

Oh. 3.46" Variation in 24 hours

=15°.42′

From 150.42', in time=1.2.48", take 3.46" and the remainder 59′.2".

Then,

24h 59.2" 24h::20 .49′52′′: 31b.43.17" true time of the moon's passing the meridian of Greenwich.

PRACTICAL EXAMPLES.

1 Required the time of the moon's culminating at Greenwich on the 14th of April 1822.

The sun's right ascension at noon (Naut. Alm.) being 1b.28'.42" and the moon's 293°.57', and on the 15th 1.32'.23" and 307°.0'.

Answer. 18h.44'.57".

3. Required the time of the moon's culminating at Greenwich on the 18th of October 1822.

The sun's right ascension at noon (Naut. Alm.) being 13.30.56", and the moon's 239°.49', and on the 19th 13h.34.41" and 252°54'.

Answer. 2.33'.37".

PROBLEM IX.

(F) Given the observed altitude of a fixed star to find its true altitude.

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