To find the true central Distance, and, hence, the Longitude of the Place of Observation :

Log. diff.19.996651

29:22:27"Log.sine 9.690648
25.40.20 Log.sine 9.636711

Half diff. of true alts. 1928 7"Log.sine 8.408737

86:48:20%Log.tan.11. 253268Log.sin. 9.999325

Longitude of the place of observ., in time=3 948: = 47°27′ east.

Example 3.

December 25th, 1825, in latitude 39:13 N.. and longitude 42:55 W., by account, at 14:4927 apparent time, the mean of several observed distances between the moon's nearest limb and the planet Mars was 94:40:22; required the longitude of the place of observation?

Observed distance of moon's nearest limb = 94:40:22%

Moon's true semi-diameter =

+15.41

Sun's right ascension at noon, December 25th,=18:15 13:
Correction of ditto for 17:417: =

Sun's reduced right ascension =
Apparent time of observation =

Right ascension of the meridian =

+ 3.16

18:1829:
14.49.27

9: 7:56:

To find the true and the apparent Altitude of the Planet Mars :-

Mars' mer. zen. dist. 48: 5:53 Nat.V.S.=332142

Natural number 395921 Log.= 5.597609

=

True altitude of Mars-15:46:46 N. co-V.S.=728063
Correction of altitude + 3.14

To find the true and the apparent Altitude of the Moon's Centre:

Moon's reduced horizontal parallax =

56:48"

Moon's reduced semi-diameter 15:28" + augmentation 13" = 15:41"

4

D's mer. zen. dist. 19:23:37"Nat. vers. sine=056740

Natural number = 108123 Log.=5.033920

D's true cent. alt.=56:37:48" Nat. co-V. S.= 164863
Correction of alt.-31. 7

Moon's ap. alt.= 56: 6:41"

To find the true central Distance, and, hence, the Longitude of the Place of Observation :

Half sum of true alts.=36:12:17"Log.co-sin. 9. 906826

December 30th, 1825, in latitude 46:30: S., and longitude 84: 15 E., by account, at 21:1015: apparent time, the mean of several observed distances between the nearest limbs of the sun and moon was 107:2:7", and, at the same time, the mean of an equal number of altitudes of the moon's upper limb was 15:40:24"; but, for want of the necessary assistants, the sun's altitude could not be taken; the height of the eye above the level of the horizon was 18 feet; required the true longitude of the place of observation?

To find the true and the apparent Altitude of the Sun's Centre:

Sun's mer. zenith distance=23:21:46 N.V. S. 081988

Natural number=165835 Log.=5.219675

To find the true central Distance, and, hence, the Longitude of the

O's ap. alt. 48:47:29%

D's ap. alt. 15. 20. 15

64: 7:44
33.27.14

Place of Observation :

sup.

O's true central alt.48:46:46"

D's true central alt. 16. 13. 31

sup.

N.V.S} 1.436349 Sum64:59:59" N.V.S.3.1.422622
N.V. S. .165671 Diff.32. 33. 15 N.V.S.. 157117

Sum = Diff. =

Sum of app. alts. 64: 7:44"Nat. V. S.=1.436349
App. cent. dist. 108:34:30" Nat.V. S. 1.318546

Sum of true alts. 64:59:59"N.V.S.sup.1. 422622

True cent. dist.=107°46′34′′ | N.V.sinel. 305299

Longitude of the place of obs., in time 53658 84:14:30% east.

Remark.-In Problem XXIX, page 320, of "The Young Navigator's Guide to the Sidereal and Planetary Parts of Nautical Astronomy," there is an interesting method given for reducing the apparent central distance between the moon and sun, or a fixed star, to the true central distance, by an instrumental operation; it being a correct mechanical mode of working the lunar observations by Gunter's scale and a pair of compasses.

PROBLEM XI.

To find the Longitude of a Place by the Eclipses of Jupiter's Satellites.

FIRST,

To know if an Eclipse will be visible at a given Place,

Convert the mean time of the eclipse at Greenwich (as given in page III. of the month in the Nautical Almanac,) into apparent time, by Problem II., page 416; and let this time be reduced to the meridian of the place of observation, by Problem IV., page 297.