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3. A ship sails on the following courses, SE. 40 miles, NE. 28 miles, SW6W. 52 miles, NWW. 30 miles, SSE. 36 miles, SEBE, 58 miles. Required her course, distance, difference of latitude and departure made good.

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It has already been observed, that sailing on the principle of the plane chart is too erroneous, , when applied to: the surface of a sphere, to be used in any. but small distances, or between the tropics, where the meridians have but little convergency, and the rhumb lines do not widely differ from portions of great circles. In all sea reckonings, the principles of the plane chart are supposed to be exact enough for the distance of a day's run ; and, at the end of

every 24 hours, the ship's place is determined in latitude and longitude, by applying the difference of latitude, found by these principles, to the latitude, and the difference of longitude to the longitude of the place, from which the ship sails. This difference of longitude is found, either by parallel sailing, middle latitude sailing, or Mercator's sailing.

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PARALLEL SAILING is the method of finding the differ. ence of longitude made by a ship, when her course and distance, on a known parailel of latitude, are given ; and the contrary.

The computations in parallel sailing depend on the following

THEOREM.

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On the globe, the difference of longitude between any two meridians is to the distance between those meridians, mčasured on any parallel of latitude, as radius to the com side of the latit.de.*

1

CORE

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* DEMONSTRATION. Let PCQ be a quadrant P. of the meridian, All the radius of the equator, BC the radius of a parallel of latitude. Then, if the quad. rant PCR be turned round I its axis Pal, the points and C will describe the cir cumferences of two circles, A which will be to each other id the ratio of their radii AQ, BC.

But AQ, the radius of the equator, is the sine of the are PQ, a quadrant, and BC, the radius of the parallel of latitude, is the sine of the arc PC, the complement of the latitude of the

COR. 1: The length of a degree of longitude in any parallel is as the cosine of the latitude,

COR. 2.

As the cosine of any parallel is to the cosine of any other parallel, so is the length of any arc of the former to the length of the corresponding arc of the latter.

Cor. 3. From this theorem is derived the method of COR.

; constructing the line of longitude, in the Ilane Scale.*. See Plane Scale.]

EXAMPLES.

1. A slip in latitude 44° 12' N. sails E. 79 miles. Reş quired her difference of longitude.

Construction.

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place C. Therefore, as the circumference of the equator : the circumference of the parallel

:: radius the cosine of tbe latitude.

But similar parts are as the wholes. Therefore, the difference of longitude between any two meridians is to the distance be. tween those meridians, measured on any parallel of latitude, as radius to the cosine of the latitude. Which was to be proved.

* The truth of the process may be easily shewn. For, since AF, equal to radius CE, is to the cosines' A50, A40, A3, &c. as 60 to 50, 40, 30, &c. respectively, an equatorial degree, or 65 miles, must have the same ratios to the degrees of the par. allels, of whose latitudes the lines are the cosines. But Eso, E40, E30, &c. are the chords of those latitudes. Therefore, the latitude being taken on the chord line, the corresponding Tumber on the line of longitude shews the length of a degree on that parallel, in both parts of an equatorial degree, or in nautical miles.

i Construction.

B

Make AC cosine 44° 12', and AB radius or sine 90° From the centre. A describe the arcs, CD, BE, passing through CB. On the arc CD, from C set off 79 miles as a .. chord ; and through D, its

А other, extremity, draw AD, prolonged till it meet the arc BE in E. Join BE, and the right line BE is the dif. ference of longitude.

This construction is a geometrical solution of the proportion in the theorem, taken in an inverted order. That is, in the similar triangles ACD, ABE, cosine latitude AC: radius AB :: departure, or meridian distance CD : difference of longitude BE.

Another Construction,

Make an angle CAB of as many degrees and mine utęs as the latitude. From the angular point A set off AC=79 miles = the departure or meridional distance. From C erect the perpendicular ÇB, meeting A AB in B. AB will be the difference of longitude required.

This construction exiribits the same proportion as the foregoing. For, in the right-angled triangle ABC, cosine latitude" : 'radius :: departure AC : difference of longitude AB.

Computation.

Computation.
Cosine latitude 45° 48'
: Radius
:: Departure 79

9885546 10 1.89763

: Difference of longitude 110

Or 1° 50'.

2'04217

Solution by the Line of Longitude.

Opposite to 44 = the latitude on the line of chords, stands 43 on the line of longitude, which is, therefore, the number of miles in a degree of longitude, in that latitude. Whence 43

60 :: 79 : 110 miles, or the difference of longitude.

Solution by the Traverse Table.

With the colatitude as a course, and the meridional dis-, tance as departure, find the corresponding distance. This distance is the difference of longitude. That is, with 46° at the bottom of the page, seek 79 in the column of departure, and the distance 1 10 is the difference of longitude.

The reason of this operation is clear from considering, that in the right-angled triangle ABC, the angle B, or the colatitude, is opposed to the departure AC, exactly the same as if it were an angle of course, and that the difference of longitude AB is the side opposite to the right angle, in like manner as the distance in plane sailing.

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2. A ship in latitude 46° 5' N. is bound to Gibraltar, which lies on the same parallel, and in 4° 46' W. longitude. By, an eclipse of the moon, she finds her present

. longitude to be 13° 8' W. What is her meridional distance ?

Construction.

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