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lunation,) it is evident that she must attract the waters of the ocean with very unequal forces: but the sun's distance from the earth being so very immense, that, compared with it, the diameter of this planet is rendered nearly insensible, bis attraction is consequently more uniform, and therefore it affects the different parts of the ocean with nearly an equal force.

By the combined effect of these two forces, the tides come on sooner when the moon is in her first and third quarters, and later when in the second and fourth quarters, than they would do if raised by the sole lunar agency: it is, therefore, the mean quantity of this acceleration and retardation that is contained in the present Table, the arguments of which are, the apparent times of the moon's reduced transit; answering to which, in the adjoining column, stands a correction, which, being applied to the apparent time of the moon's passage over the meridian of any given place by addition or subtraction, according to its title, the sum, or difference, will be the corrected time of transit. Now, to the corrected time of transit, thus found, let the time of high water on full and change days, at any given place in Table LVI., be applied by addition, and the sum will be the time of high water at that place, reckoning from the noon of the given day: should the sum exceed 12:24", or 24:48, subtract one of those quantities from it, and the remainder will be the time of high water very near the truth.

Example 1. Required the time of high water at Cape Florida, America, March 7th, 1824 ; the longitude being 80:5! west, and the time of high water on full and change days 7:30" ? Moon's transit over the meridian of Greenwich, per Nautical Almanac, March 7th, 1824, is

5. 20:
Correction from Table XXXVIII., answering to retardation of
transit 58, and longitude 80:5! west =

+ 12.23
Moon's transit reduced to the meridian of Cape Florida 5:14 23:
Correction answering to reduced transit (5?14"23) in Table
XXXIX., is

1. 5. 0 Corrected time of transit

49:23: Time of high water at Cape Florida on full and change days 7.30. 0 Time of high water at Cape Florida on the given day = 11.39723!

Example 2.
Required the time of high water in Queen Charlotte's Sound, New.
Zealand, April 13th, 1824; the longitude being 174:56: east, and the
time of high water on full and change days 9:07?

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Moon's transit over the meridian of Greenwich, per Nautical

Almanac, April 13th, 1824, is
Correction from Table XXXVIII., answering to retardation of

transit 50., and longitude 174:56! west =

10:275 0:

23, 29

.

Moon's transit reduced to the meridian of Queen Charlotte's
Sound

10: 3:31: Correction answering to reduced time of transit (10.3.31) in Table XXXIX., is

+ 23. 0

Corrected time of transit
Time of high water at given place on full and change days

10:26:31: 9. 0.0

Time of high water at Queen Charlotte's Sound, past noon of the given day

19:26:31: Subtract 12. 24. 0

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Reduction of the Moon's Horizontal Parallax on account of the

Spheroidal Figure of the Earth.

Since the moon's equatorial horizontal parallax, given in the Nautical Almanac, is determined on spherical principles, a correction becomes necessary to be applied thereto, in places distant from the equator, in order to reduce it to the spheroidal principles, on the assumption that the polar axis of the earth is to its equatorial in the ratio of 299 to 300; and, when very great accuracy is required, this correction ought to be attended to, since it may produce an error of seven or eight seconds in the computed lunar distance. The correction, thus depending on the spheroidal figure of the earth, is contained in this Table; the arguments of which are, the moon's horizontal parallax at the top, and the latitude in the left-hand column; in the angle of meeting will be found a correction, expressed in seconds, which being subtracted from the horizontal parallax given in the Nautical Almanac, will leave the horizontal parallax agreeably to the spheroidal hypothesis.

Thus, if the moon's horizontal parallax, in the Nautical Almanac, be 57:58", and the latitude 51:48' ; the corresponding correction will be 7 seconds subtractive. Hence the moon's horizontal parallax on the spheroidal hypothesis, in the given latitude, is 57:51".

Remark.-The corrections contained in this Table may be computed by the following

Rule.

To the logarithm of the moon's equatorial horizontal parallax, reduced to seconds, add twice the log. sine of the latitude, and the constant log, 7.522879;* the sum, rejecting the tens from the index, will be the logarithin of the corresponding reduction of parallax.

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Let the moon's horizontal parallax be 57.58”, and the latitude 51:48!; required the reduction of parallax agreeably to the spheroidal hypothesis ?

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Moon's equatorial horiz. par. 57:58. = 3478". Log. = 3.541330
Latitude

51:48: Twice the log. sine=19.790688
Constant log

7.522879

Reduction of horizontal parallax =

7". 159

Log.=0.854897

TABLE XLI.

Reduction of Latitude on account of the Spheroidal Figure of the Earth.

Since the figure of the earth is that of an oblate spheroid, the latitude of a place, as deduced directly from celestial observation, agreeably to the spherical hypothesis, must be greater than the true latitude expressed by the angle, at the earth's centre, contained between the equatorial radius and a line joining the centre of the earth and the place of observation. This excess, which is extended to every second degree of latitude from the equator to the poles, is contained in the present Table; and which, being subtracted from the latitude of any given place, will reduce that latitude to what it would be on the spheroidal hypothesis : thus, if the latitude be 50 degrees, the corresponding reduction will be 11:42", subtractive ; which, therefore, gives 49:48:18". for the reduced or spheroidal latitude.

Remark.-The corrections contained in this Table may be computed by the following rule; viz.,

To the constant log. .003003,+ add the log. co-tangent of the latitude,

* The arithmetical complement of the log. of the earth's ellipticity assumed at góg.

+ The excess of the spherical above the elliptic arch in the parallel of 45 degrees from the equator, is 11'.887, or ll’53" (Robertson's Navigation, Book VIII., Article 134): hence 450 – 11'53" = 44° 48' 7", the log. co-tangent of which, rejecting the index, is .003003.

and the sum will be the log. co-tangent of an arch; the difference between which and the given latitude will be the required reduction.

Example Let it be required to reduce the spherical latitude 50:48! to what it would be if determined on the spheroidal principles; and, hence, to find the reduction of that latitude.

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A General Traverse Table ; or Difference of Latitude and Departure. This Table, so exceedingly useful in the art of navigation, is drawn up

in a manner quite different from those that are given, under the same denomination, in the generality of nautical books; and, although it occupies but 38 pages, yet it is more extensive than the two combined Tables of 61 pages, which are contained in those books. In this Traverse Table, every page exhibits all the angles that a ship's course can make with the meridian, expressed both in points and degrees; which does away with the necessity of consulting two Tables in finding the difference of latitude and the departure corresponding to any given course and distance. If the course be under 4 points, or 45 degrees, it will be found in the left-hand compartment of each page; but that above 4 points, or 45 degrees, in the right-hand compartment of the page. The distance is given, in numerical order, at the top and bottom of the page, from unity, or 1, to 304 miles ; which comprehends all the probable limits of a ship's run in 24 hours; and, by this arrangement, the mariner is spared the trouble of turning over and consulting twenty-three additional pages. Although the manner of using this Table must appear obvious at first sight, yet since its mode of arrangement differs so very considerably from the Tables with which the reader may have been hitherto acquainted, the following Problems are given for its illustration.

PROBLEM I.

Given the Course and Distance sailed, or between two Places, to find the Difference of Latitude and the Departure.

Rule. Enter the Table with the course in the left or right-hand column, and the distance at the top or bottom; opposite to the former, and under or over the latter, will be found the corresponding difference of latitude and departure : these are to be taken out as marked at the top of the respective columns if the course be under 4 points or 45 degrees, but as marked at the bottom if the course be more than either of those quantities.

Note. If the distance exceed the limits of the Table, an aliquot part thereof may be taken, as a half, third, fourth, &c.; then the difference of latitude and departure corresponding to this and the given course, being multiplied by 2, 3, 4, &c., (that is, the figure by which such aliquot part was found,) the product will be the difference of latitude and departure answering to the given course and distance.

Example 1. A ship sails S.S.W. W. 176 miles ; required the difference of latitude and the departure?

Opposite 24 points and under 176 miles, stand 155. 2 and 83.0: hence the difference of latitude is 155. 2, and the departure 83.0 miles.

Example 2. A ship sails N. 57. E. 236 miles ; required the difference of latitude and the departure ?

Opposite to 57, and under 236 miles, stand 128.5 and 197. 9: hence the difference of latitude is 128.5, and the departure 197.9 miles.

Example 3. The course between two places is E. b. S. S., and the distance 540 miles ; required the difference of latitude and the departure ?

Distance divided by 2, gives 270 miles ; under or over which, and opposite to 64 points, stand

91.0 and 254. 2 Multiply by

2

2

Products =

182.0 and 508.4: hence the difference of latitude is 182.0, and the departure 508.4 miles,

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