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

Answer. 125.42'.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. 126.38'.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, 176.42.55", and on the 19th, 176.47' 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 76.23'.13" (Tab.VIII.); O's right ascension (Naut. Alm.) December 1st=16h.28'.16", and on the 2d=169.32.35'.

Answer. 14.9.52.16".

PROBLEM VIH.

(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'st 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'.1apparent 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.

of 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 tbe 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 płanet.

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

D's right ascension at noon 13th August 1822 =940.59ʻ =. 19'.56"
O's right ascension at noon 13th August 1822

=95.30', 4'

Time of culminating rearly

204.49'.521

O's right ascen. Aug. 13th = 9h30. 4". D's right ascension Aug. 13th= 940.59 O's right ascen. Aug. 14th =9*33*.50'|'s right ascension Aug. 14th=1109.41'

Variation in 24 hours

=Oh. 3.46|Variation in 24 hours

= 150.42

From 15°42', in time=1h 2.48", take 3.'46" and the remainder-59.2". Then,

24h - 59.2" : 245 :: 20.49'52" : 316.43.17" 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 16.28'.42" and the moon's 2930.57', and on the 15th 15.32.23" and 3070.0'.

Answer. 186.44'.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 13h.30'.56", and the moon's 2390.49', and on the 19th 13h.34.41" and 2520.54'.

Answer. 26.33.37".

PROBLEM IX.

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

RULE. From the observed altitude, subtract the refraction (Table 1V.) 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'
Refraction, Table IV.

-2.32"

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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,t 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(Table IV.), and then add the parallax in altitude g (Table VI.). If the altitude be taken at sea, the dip of the

* 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 ex. tremity of the vertical diameter. IS. 87. et seq.

§ S. 95.

horizon (Table V.) must be deducted, the last result will be the true altitude of the sun's centre.

EXAMPLE I,

On the 13th of March 1822, if the altitude of the sun's lower limb, observed at sea, be 18°.40ʻ; required the true central altitude, the height of the eye being 22 feet.

Observed altitude of the O's lower limb=189.40
O's semidiameter, Naut. Alm.

+ 16'. 6"

180.56'. 6" Refraction for 19° of altitude (Table IV.)= –2.44"

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Or, (16'.6" +81)-(2'.44" +4'.28"')=9'.2" and 18°.40' +9'.2" =180.49'.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? Answer. 28o.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 applicd to the observed altitude, previous to the other corrections.

To the cosine of the moon's apparent altitude, add the logarithm of the horizontal parallax in seconds (found above), the sum, rejecting 10 from the index, will be the logarithm of the moon's parallax in altitude in seconds (T. 96.), from which take the refraction of the moon in altitude (Table IV.), and the remainder will be the moon's correction, which added to the apparent altitude will give the true altitude.

EXAMPLE I.

On the 18th of July 1822, in longitude 113°.10' W. at 206.45':30", if the observed altitude of the moon's lower limb be 45°22'.3", error of the quadrant 58" subtractive, and the eye 21 feet above the level of the sea; what is the true central altitude ? Time at the given place

204.45'.30" Longitude 113°.10 w. in time = 76.32'.40% W.

28h.18'.10"
24

Time at Greenwich on the 19th of July= 45.18'.10"

D's semidiameter at noon, 19th=16'.9", Horizontal parallax
D's semidiameter at midnight, 19th=16'.3', Horizontal parallax

59'.15" 58.54"

Difference 0.6"

Difference 0.21"

12h : 6"::45.18'.10":0. 2

12h : 21"::4h. 18'.10': 7" »'s semidiameter at noon 16'. 9" Horizontal parallax at noon

59'.15" D's semidiam. at 45.18'.10" =16'. 7|Horizontal paral. at 4". 18'.10= 59' 8' D'saugmentation (TableVII.) = 11"

60

D's true semidiameter

= 16'.18"

In seconds

=3548

D's observed altitude = 45°.22. 3"||Cos') 's app. alt. 450.33.1"=9.84911 Error of the quadrant

-58" Horizontal parallax 3548" log=3•54998

450.21'. 5" ||Parallax in altitude 2507" log=8.39909 Semidiameter

+ 16'.18"
Then 2507"

41.47" 450.37'. 23" ||'s refraction

=-0.56" Dip (Table V.)

-4'.22'
D's correction

40'.51" App. altitudo ) 's centre = 450.33.1"

Lastly, 450.33.1" +40'.51" = 46°.13'.52", true altitude of the moon's centre.

2. On the 22d of April 1822, in longitude 105° E. of Greenwich, at 135.30', if the observed altitude of the moon's upper

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