traction to the geocentric right ascension and declination of the planet at the noon preceding the Greenwich time, according as these elements may be increasing or decreasing; the result will be the correct right ascension and declination of the given planet.

Example.

Required the geocentric right ascension and declination of the planet Venus, January 3rd, 1836, at 5:20:36 mean time; the longitude being 150:40:15" east of Greenwich?

Given mean time

.52036:

Longitude 150:40:15% east, in time -10. 2.41

The difference of R. A. between the noons of Jan. 2nd and 3rd is

Venus's R. A. at noon, January 2nd= 20:22′′58:

Venus's declination at noon, January 2nd 21: 1 4" south.

In the same manner may the heliocentric longitude and latitude of a planet be reduced to any given time under a known meridian.

Remarks.-1. In the Nautical Almanac, between pages 267 and 358, the geocentric right ascensions of the planets are given to hundredths of a second, and their declinations to tenths of a second; but, since the nearest second in either of these elements will always be sufficiently exact for the ordinary purposes of navigation; therefore, the decimals may be dispensed with, as in the above example:-see the note which is given at the end of the solution to Problem XIV., in page 361.

2. It is only the four bright planets, viz. Venus, Mars, Jupiter, and Saturn, that concern the practical navigator. The other planets, viz., Mercury, the Georgian, and the asteroids, Vesta, Juno, Pallas, and Ceres, are but very rarely visible to the naked eye; the first, because its proximity to the sun causes it to be generally lost in the splendour of the solar rays; the second, because of its immense distance from the earth; and the asteroids, because of their comparative minuteness; and therefore they are of no manner of use whatever for nautical purposes on the high seas.

PROBLEM XVIII.

Given the observed Mean Time, per Watch, of the Sun's Transit over the Meridian of a Given Place; to find the correct Mean Time of Transit.

Since the time of the sun's transit over the meridian of any place is indicated by the instant that a correct sun-dial projects its shadow along the plane of the meridian, either north or south, according to its position with respect to the equator, viz., when the shadow shows 0:00:; therefore the solution of this problem is simply to convert apparent noon into mean noon: this may be done by the following

RULE.

To the given observed mean time of transit apply the longitude in time, as directed in Problem III., page 342; the result will be the corresponding mean time at Greenwich: to which, let the equation of time, in page II. of the month in the Nautical Almanac, be reduced by Problem XIV., page 357. Now, when the equation of time is noted for addition in page I. of the month, (the sign in this page being always the reverse of what it is in page II.), let the apparent noon be called 0:0:0:; but, when it is noted for subtraction, call the apparent noon 24:00: :-Then, in the first case, 0:00: + the reduced equation of time; and, in the second, 24:0:0: - the reduced equation of time will

be the correct mean time of the sun's transit or passage over the meridian of any given place.

Example.

September 29th, 1836, at 23:50:7: mean time, per watch, the sun was observed to pass the meridian of a place 90:45 west of Greenwich; required the correct mean time of transit ?

Equation of time at noon, Sept. 30 = 10" 5:52

Reduced equation of time

10" 10:24.-Now, since the equa

tion of time is marked subtractive in page I. of the month in the Ephemeris; therefore, 24-10:10:24 23:49:49:76, is the correct mean time of the sun's transit over the given meridian.

Note. Since this is merely the conversion of apparent noon into mean noon; it does not require any further illustration.

See Explanatory Articles 35 and 36, relative to the Equation of Time, between pages 310 and 313.

PROBLEM XIX.

Given the Mean Time at Ship, per Watch, and the Sun's Horary Distance from the Meridian; to find the Correct Mean Time :-or, to convert Apparent Time into Mean Time.

RULE.

To the given mean time at ship, or place, apply the longitude, in time, as directed in Problem III., page 342; the result will be the corresponding mean time at Greenwich:-to which let the equation of time, in page II. of the month in the Nautical Almanac, be reduced by Problem XIV., page 357. Now, this reduced equation being applied to the sun's horary distance, viz., the apparent time, with a contrary sign to what it is marked in page II.; or, which may be something

BB

plainer, according to its sign in page I. of the month; the result will be the correct mean time at the given place.

Example.

February 1st, 1836, at 3:10:20: mean time, per watch, in longitude 50:40. west, the sun's horary distance from the meridian, or the apparent time, was 3:6:35; required the correct mean time?

Prop. log. 3. 1451

Equation of time at noon, Feb. 1st +3 34:91

=

3:10:17:66, as required.

Remarks-1. Although the above example is worked out to the hundredths of a second; yet, it is not absolutely necessary to do so: because, for all practical purposes at sea, the nearest second will be found quite sufficient; as pointed out in the note which is appended to Problem XIV., in page 357:-This will appear manifest by reflecting that an error of one second in the time will only affect the longitude to the value of a quarter of a mile; which, in nautical operations, on the ocean, may well pass unnoticed, as not worthy of a consideration.

2. The young navigator must bear in mind that apparent time only relates to the sun; it has nothing whatever to do with the rest of the heavenly bodies. See Articles 35 and 36, between pages 310 and 313, relative to the Equation of Time, &c. &c.

PROBLEM XX.

Given the Mean Time, and the Horary Distance of the Moon, a Fixed Star, or a Planet, from the Meridian; to find the Correct Mean Time.

RULE.

To the given mean time at ship, or place, apply the longitude, in time, as directed 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," which is given in page II. of the month in the Nautical Almanac), be reduced by Problem V., page 344.- Reduce the star's right ascension, as given in the pages between 368 and 407 of the Nautical Almanac, to the given day; but, the right ascension of the moon, or a planet, must be reduced to the Greenwich time by Problem XVI., page 364, or XVII., page 366 :Then, if the given celestial object be west of the meridian, let its horary distance be applied by addition; but if east by subtraction to its corrected right ascension:-the result will be the right ascension of the meridian.-From this, increased by 24 hours if necessary, subtract the reduced mean sun's right ascension; and the remainder will be the correct mean time.

Example 1.

May 1st, 1836, at 10:21 24 mean time, per watch, in longitude 70:36 east, the horary distance of Antares, east of the meridian, was 3:2013; required the correct mean time?

Given mean time = . 10:21 24: | Mean sun's R. A. at

May 3rd, 1836, at 17:4437: mean time, per watch, in longitude 76:42 east, the horary distance of the moon, west of the meridian, was 31233; required the correct mean time?

Given mean time = 17:44:37! || Mean sun's R. A. at

Longitude 76:42 east,

noon, May 3rd = Equation to 12:37:49:

2:45 39:31

+2. 4.49

Mean sun's cor. R. A.=2:47:43:80