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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 2 The right ascension of Arcturus being 14h.7'.32".t "'s right ascension (1822)+24h - - =38b. 7.32". G)'s right ascension, December 1st, 1822 - = 16*.28'.16".

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Go’s right ascension, December 1st, 1822, at noon= 16*.28.16". G)'s right ascension, December 2d, 1822, at noon= 16*.32.35".

Increase of the G)'s right ascension in 24 hours = Oh. 4'.19".

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I. At what time will Aldebaran culminate at Greenwich on the 20th November 1822.2 Therightascension of Aldebaran being4".25'43"(Tab. VIII.) the sun's right ascension at noon (Naut. Alm.) 15h.41.29" on November 20th, and on the 21st 15b.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 9.58".58" (Tab.VIII.); G)'s right ascension Feb. 6th (Naut. Alm.) being 21". 18.40", and on the 7th 21.h.22.40". Answer. 12h.38".6”. 3. At what time, on the 18th of December 1822, will Sirius, the Dog Star, culminate at Greenwich 2 The right ascension of Sirius being 6.87°. 18" (Tab. VIII.); G)'s right ascension 18th December, (Naut. Alm.) at noon, 17*.42.55", and on the 19th, 17b.47".21”. Answer. 12h.52'. 4. At what time, on the 1st of December 1822, will Castor culminate at Greenwich 2 The right ascension of Castor being 7.23.13" (Tab.VIII.); G)'s right ascension (Naut. Alm.) December 1st= 16*.28'.16", and on the 2d=16h.32.35". Answer. 14h.52'.16".

* 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 24* : 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 exeeed two minutes of time.

PROBLEM VIII.

(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 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.

* 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 12*.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 12%. 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.

EXAMPLE.

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

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Time of culminating rearly - - - - 20%. 49'.52”

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Variation in 24 hours =Oh. 5,46, Variation in 24 hours - = 15°.42’

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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 1.h.28/.42" and the moon’s 293°.57, and on the 15th 1.32.23" and 30.7°.O.

Answer. 18".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 135.30.56", and the moon’s 239°.49', and on the 19th 13h.34/.41’’ and 25.2°.54'.

Answer. 2b, 33.37".

PROBLEM IX.

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

RULE. From the observed altitude, subtract the refraction (Table IV.) If the star be observed at sea, subtract the dip of the horizon (Table V.) See R. 94.

EXAMPLE I.

The observed altitude of Spica Virginis was 20°.39.40°, the error" of the quadrant 19" subtractive, and the height of the eye 18 feet above the level of the sea; required the correct altitude. Observed altitude of Spica Virginis 20°.39'.40"

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Or, 20°.39.40"—(2.32"-- 4.3"+ 19")=20°.32.46". Answer.

2. Suppose the observed altitude of Regulus to be 45°.13%. 15", the height of the eye 14 feet above the level of the sea, and the error of the quadrant 5.6" additive, required the true altitude.

Answer. 45°. 13.50".

PROBLEM X.

(G) Given the observed altitude of the sun's lower or upper limb,f to find the true altitude of its centre.

RULE. To the observed altitude apply the semidiameter (taken from page III. of the month in the Nautical Almanac) by addition or subtraction, according as the lower or upper limb has been observed; from this result subtract the refraction f (Table IV.), and then add the parallax in altitude § (Table VI.). If the altitude be taken at sea, the dip of the horizon (Table V.) must be deducted, the last result will be the true altitude of the sun's centre.

* Observations taken with a quadrant are liable to errors, arising from the bending and elasticity of the index, and the resistance it meets with in turning round its centre. These errors, though they cannot in all cases be avoided, may be pretty accurately allowed for by a correct observer.

+ The sun's upper limb is the upper edge of its face, or the uppermost extremity of the vertical diameter; and the lower limb is the lower edge, or the lower extremity of the vertical diameter.

- : S. 87. et seq. § S. 95.

EXAMPLE I.

On the 13th of March 1822, if the altitude of the sun's lower limb, observed at sea, be 18°.40'; required the true central altitude, the height of the eye being 22 feet. Observed altitude of the G’s lower limb- 18°40' G)'s semidiameter, Naut. Alm. - - = + 16. 6."

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Or, (16.67 +8")–(2.44"+4.28")=9'.2" and 18°40' +9'.2" = 18°.49'.2". Answer. Note. The index error, if any, of the quadrant must be applied to the observed altitude, previous to the other corrections. 2. On the 3d of September 1822, suppose at sea the altitude of the sun's upper limb to be 28°.31'.30”, the index error 40" additive, height of the eye 12 feet, and the sun's semidiameter 15.54"; what is the true altitude of the sun's centre 2 Answer. 28°.11". 19'.

PROBLEM XI.

(H) Given the observed altitude of the moon's lower or upper limb, to find the true altitude of its centre.

RULE. Find the moon's semidiameter and horizontal parallax for the time and place of observation (C. 271.), and increase the semidiameter by the augmentation answering to the moon's altitude (Table VII.).

To the observed altitude" apply the augmented semidiameter by addition or subtraction, according as the lower or upper limb has been observed, and, if the observation has been made at sea, subtract the dip of the horizon (Table V.); the result will be the apparent altitude of the moon’s centre.

* Here, as in the preceding problem, the index error of the quadrant, if any, must be applied to the observed altitude, previous to the other corrections.

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