Given the Latitude of a Place, the Altitude, Right Ascension, and Declination of a known fixed Star, and the Sun's Right Ascension ; to find the apparent Time, and, hence, the Error of the Watch.

RULE.

Find the true altitude of the star, by Problem XVII., page 327; and let its right ascension and declination, as given in Table XLIV., be reduced to the night of observation; then,

With the latitude of the place, the star's true altitude, and its reduced declination, compute its horary distance from the meridian, by any of the methods given in the last problem.

Now, if the star be observed in the western hemisphere, let its meridian distance, thus found, be added to its reduced right ascension, but, if in the eastern hemisphere, subtracted from it, and the sum or remainder will be the right ascension of the meridian; from which, (increased by 24 hours, if necessary,) subtract the sun's right ascension at noon of the given day, and the remainder will be the approximate time of observation. Reduce this to Greenwich time, by Problem III., page 297, and find the proportional part of the variation of the sun's right ascension, for the given day, answering thereto and 24 hours, by Problem XII., page 317; . which, being subtracted from the approximate, will give the apparent time of observation: hence the error of the watch may be known.

Note. For the principles of this rule, see "The Young Navigator's Guide," page 156.

Example 1.

January 1st, 1825, in latitude 40:29. N., and longitude 59:45: W., the mean of several altitudes of a Arietis, west of the meridian, was 36:29:48", that of the corresponding times, per watch, 11:929, and the height of the eye above the level of the sea 19 feet; required the apparent time, and the error of the watch?

Reduced R. A. of a Arietis=157" 19, and its reduced dec,=22°37′50′′N.

Observed alt. of a Arietis=36:29'48"; hence, its true alt. is 36:24:20%

Star's horary dist., west of the merid. 4 245 Log.rising 5.70795.2

=

+ 3.59. 0

15:11 45

Correction of approximate time, ans. to Greenwich time

15:1145, and variation of sun's R. A. 4" 24. 8,

Apparent time of observation =

Time of observation, per watch, =

Watch slow for apparent time =

028:

Example 2.

January 1st, 1825, in latitude 39°20′30′′ S., and longitude 75:40. E., the mean of several altitudes of Procyon, east of the meridian, was 27° 15:47", that of the corresponding times, per watch, 9:3023, and the height of the eye above the level of the sea 19 feet; required the apparent time of observation, and the error of the watch?

Procyon's reduced R. A. 7:30 8!, and its reduced dec. 5:39:58 N.

Observed alt, of Procyon=27:15.47%, hence, its true alt. is=27:9′46′′

Procyon's true alt. 27. 9.46 Nat.co-V.S.=543480

Procyon's horary dist., east of the mer.=31020: Log.rising=5.51252.7

Correction of approximate time, answering to Greenwich time 42949, and variation of sun's right ascension 4" 24′.8=

. 9.30.23

116:

Note. When the star's horary distance east of the meridian exceeds the right ascension, the latter is to be increased by 24 hours, in order to find the right ascension of the meridian.

In finding the error of a watch by sidereal observation, two or more stars should be observed, and the error of the watch deduced from each star separately. And, if an equal number of stars be observed on different sides of the meridian, and nearly equidistant therefrom, it will conduce to still greater accuracy; because, then, the errors of the instrument and the unavoidable errors of observation will have a mutual tendency to correct each other. The mean of the errors, thus deduced, should be taken for the absolute error of the watch.

PROBLEM V.

Given the Latitude and Longitude of a Place, and the Altitude of a Planet, to find the Apparent Time of Observation.

RULE.

Reduce the estimated time of observation to the meridian of Greenwich, by Problem III., page 297; to which time let the planet's right ascension and declination be reduced, by Problem VII., page 307; and let the sun's right ascension, at noon of the given day, be also reduced to that time, by Problem V., page 298. Reduce the observed central altitude of the planet to its true central altitude, by Problem XVI., page 325.

Then, with the latitude of the place, the planet's reduced declination, and its true central altitude, compute its horary distance from the meridian, by any of the methods given in Problem III., pages 384 to 392. Let the planet's horary distance from the meridian, thus found, be applied to its reduced right ascension, by addition or subtraction, according as it may be observed in the western or in the eastern hemisphere, and the right ascension of the meridian will be obtained; from which (increased by 24 hours, if necessary,) subtract the sun's reduced right ascension, and the remainder will be the apparent time of observation.

Note. When the planet's horary distance east of the meridian exceeds its right ascension, the latter is to be increased by 24 hours, in order to find the right ascension of the meridian.

Example 1.

January 3d, 1825, in latitude 50:30: N., and longitude 48:45. W., the mean of several altitudes of Jupiter's centre, east of the meridian, was 23:41:55%, that of the corresponding times, per watch, 91, and the height of the eye above the level of the sea 16 feet; required the apparent time of observation?

Obs. cent.alt. of Jupiter=23:41:55%; hence, its true cent. alt. is 23:35:56′′

January 16th, 1825, in latitude 34:45 S., and longitude 80:30? E., the mean of several altitudes of Venus' centre, west of the meridian, was 22:53:25%, that of the corresponding times, per watch, 7:20745, and the height of the eye above the level of the sea 18 feet required the apparent time of observation?

Sun's right ascension at noon, January 16th, 19:52:41:

Correction of ditto for 158"45! =

Sun's reduced right ascension =

+ 0.21