To the mean sun's reduced right ascension let the given mean timebe added; and the sum, abating 24 hours, if necessary, will be the right ascension of the meridian (Problem VI., page 345); the difference between which and the moon's reduced right ascension will be her horary distance from the meridian. If the latitude of the ship and the moon's declination are of contrary names, let their sum be taken; but, if of the same name, their difference; the result will be her meridional zenith distance.-Then, To the log. rising answering to the moon's horary distance from the meridian, add the log. co-sines of her declination, and the latitude of the ship; the sum, abating 20 in the index, will be the logarithm of a natural number; which, being added to the natural versed sine of the moon's meridional zenith distance, will give the natural co-versed sine of her true central altitude.-From the true altitude of the moon, thus found, subtract the correction corresponding thereto, and her reduced horizontal parallax, in Table XIX., and the remainder will be her apparent central altitude. Example. Required the true, and the apparent altitude of the moon's centre, January 7th, 1836, at 11 24 25 mean time, in latitude 50: 10: north, and longitude 60: west of the meridian of Greenwich? D's merid. zen. dist. 35°15' 7"Nat. ver. sine 183378 True alt. of D's centre=26:38:23 N.co-v.sine 551621 App. alt. of D's centre =25:49:49" as required. Note.-There is a cipher annexed to the log. rising, so as to make the number of its decimal places correspond with that of the log. cosines. PROBLEM III. Given the Latitude and Longitude of a Ship, or Place, and the Mean Time; to find the True, and the Apparent Altitude of a Planet's Centre. RULE. Reduce the given mean time to the meridian of Greenwich by Problem III., page 342:-to which let the mean sun's right ascension be reduced by Problem V., page 344: and the planet's geocentric right ascension and declination by Problem XVII, page 366. To the mean sun's reduced right ascension let the given mean time be added; and the sum, abating 24 hours if necessary, will be the right ascension of the meridian (Problem VI., page 345); the difference between which and the planet's reduced right ascension will be its horary distance from the meridian. If the latitude of the ship and the planet's declination are of contrary names, let their sum be taken; otherwise, their difference; and the planet's meridional zenith distance will be obtained.-Then, To the log. rising answering to the planet's horary distance from the meridian, add the log. co-sines of its declination, and the latitude of the ship; the sum, abating 20 in the index, will be the logarithm of a natural number; which, being added to the natural versed sine of the planet's meridional zenith distance, will give the natural co-versed sine of its true central altitude.-Now, with the planet's true altitude, thus found, enter Table XIX., and take out the equation, or correction corresponding to the reduction of a star's true altitude; the difference between which and the planet's parallax in altitude, Table VI., will leave a correction; which, being added to the true altitude, will give the apparent altitude of the planet. Example. Required the true, and the apparent altitude of Venus, January 3rd, 1836, at 7:5045 mean time, in latitude 45:34: south, and longitude 80:30 east of the meridian of Greenwich? 2 Given mean time = 7:50:45: Mean sun's R. A. at Long. 80:30: east, in time= Given mean time = 7:50:45 = R. A.= 20:28:42:Ditto, declination= 20:42:16" S. Venus's mer. zen. dis.=24:51:44" Nat. ver. sine 092678 Venus's true cen. alt. 12:45:33 Nat.co-v.sine 779148 Red., Ta. XIX.4'6" = Diff. +4:00" Venus's apparent alt.=12:49:33", as required. Note.-There is a cipher annexed to the log. rising, so as to make the number of its decimal places correspond with that of the log. cosines. PROBLEM IV. Given the Latitude and Longitude of a Ship, or Place, and the Mean Time; to find the True, and the Apparent Altitude of a fixed Star. RULE. Reduce the given mean time to the meridian of Greenwich by Problem III., page 342; to which let the mean sun's right ascension be reduced by Problem V., page 344.-Take the star's right ascension and declination from the Nautical Almanac, between pages 368 and 407 To the mean sun's reduced right ascension let the given mean time be added; and the sum, abating 24 hours, if necessary, will be the right ascension of the meridian (Problem VI., page 345); the difference G G between which and the star's right ascension will be its horary distance from the meridian. If the latitude of the ship and the star's declination are of contrary names, let their sum be taken; but if of the same name, their difference : the result will be the star's meridional zenith distance.-Then, To the log. rising answering to the star's horary distance from the meridian, add the log. co-sines of its declination, and the latitude of the ship the sum, abating 20 in the index, will be the logarithm of a natural number; which, being added to the natural versed sine of the star's meridional zenith distance, will give the natural co-versed sine of its true altitude.-Now, to the star's true altitude, thus found, let the correction corresponding thereto in Table XIX. be added; and the sum will be its apparent altitude. Example. Required the true, and the apparent altitude of the star Procyon, January 10th, 1836, at 9:31:39: mean time, in latitude 39:20:30: south, and longitude 75:40: east of Greenwich? Given mean time = 9:31 39: Mean sun's R. A. at Long. 75:40 east, time = Greenwich time = Star's mer. zen. dist. 44:58:57"Nat. ver. sine 292678 Note.-In the above Example, as in the three preceding ones, a cipher is annexed to the log. rising, so as to make the number of its decimal places correspond with the number of decimals in the log. co-sines. The reader will please to observe that the natural sines may be used in the solution of the four last problems, instead of the versed sines in this case, if the natural number be subtracted from the natural co-sine of the object's meridional zenith distance, the natural sine of its true altitude will be obtained.-Thus, in the above Example, the star's meridional zenith distance is 44:58:57" :-Now, the natural co-sine of this is 707322; from which let the natural number 184724 be subtracted, and the remainder 522598, is the natural sine of the star's true altitude; the arch corresponding to which is 31:30′24′′.The natural sines, and natural co-sines are comprehended in Table XXVII., between pages 93 and 137 of the second volume.-See the first paragraph in page 57, and the Remark in page 60, of the present volume. The four preceding problems are evidently the converse of those for finding the mean time, as given in pages 435, 437, 439, and 441. SOLUTION OF PROBLEMS RELATIVE TO THE The Longitude of a given place on the earth, is that arc or portion of the equator which is intercepted between the first or principal meridian and the meridian of the given place; and is denominated east or west, according as it may be situate with respect to the first meridian. The first or principal meridian is an imaginary great circle passing through any remarkable place and the poles of the world: hence it is entirely arbitrary; and, therefore, the British reckon their first meridian to be that which passes through the Royal Observatory at Greenwich; the French esteem their first meridian to be that which passes through the Royal Observatory at Paris; the Spaniards, that which passes through Cadiz, &c. &c. &c. Every part of the terrestrial sphere may be conceived to have a meridian line passing through it, cutting the equator at right angles: hence there may be as many different meridians as there are points in the equator. Every meridian line, with respect to the place through which it passes, may be said to divide the surface of the earth into two equal parts, called the eastern and western hemispheres. Thus, when the face of an observer is turned towards the north pole of the world, the hemisphere which lies on his right-hand is called east, and that on his |