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Required her latitude, direct course, and distance.

Ans. Latitude= 27° 23′ N.

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4. A ship from Bermuda, latitude 32° 22', sails on the fol

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Required her latitude, direct course, and distance.

Ans. Latitude= 33° 53' N.

Course N. 57° 22′ E.
Distance=168.4 miles.

(196.) When the water through which a ship is moving has a progressive motion, the ship's progress is affected in the same manner as if she had sailed in still water, with an additional course and distance equal to the direction and motion of the current.

Ex. 5. If a ship sail 125 miles N.N.E. in a current which sets W. by N. 32 miles in the same time, required her true course and distance.

Form a traverse table containing the course sailed by the ship and the progress of the current, and find the difference of latitude and departure. The resulting course and distance is found as in the preceding examples.

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Hence the course is found by plane sailing N. 7° 42′ E., and the distance=122.8 miles.

Ex. 6. A ship sails S. by E. for two hours at the rate of 9 miles an hour; then S. by W. for five hours at the rate of 8 miles an hour; and during the whole time a current sets W. by N. at the rate of two and a half miles an hour. Required the direct course and distance.

Ans. The course is S. 21° 51′ W.

Distance 57.6 miles.

PARALLEL SAILING.

(197.) Parallel sailing is when a ship sails exactly east of west, and therefore remains constantly on the same parallel of latitude. In this case the departure is equal to the distance sailed, and the difference of longitude may be found by the following

THEOREM.

The cosine of the latitude of the parallel is to radius, as the distance run is to the difference of longitude.

P

Let P be the pole of the earth, C the center, AB a portion of the equator, and DE any parallel of latitude; then will CA be the radius of the equator, and FD the radius of the parallel. Let DE be the distance sailed by the ship F on the parallel of latitude, then the difference of longitude will be measured by AB, the c arc intercepted on the equator by the meridians passing through D and E.

A

E

B

Since AB and DE correspond to the equal angles ACB, DFE, they are similar arcs, and are to each other as their radii. Hence

FD: CA:: arc DE: arc AB.

But FD is the sine of PD, or the cosine of AD, that is, the cosine of the latitude, and CA is the radius of the sphere; hence

Cosine of latitude : R:: distance: diff. longitude.

Cor. Like portions of different parallels of latitude are to cach other as the cosines of the latitudes.

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The length of a degree of longitude in different parallels may be computed by this theorem. A degree of longitude at the equator being 60 nautical miles, a degree in latitude 40° may be found by the proportion

R: cosine 40° :: 60: 45.96, the required length. The following table is computed in the same manner. (198.) Table showing the length of a degree of longitude for each degree of latitude.

Lat. Miles. Lat. Miles. Lat. Miles. Lat. Miles. Lat. Miles. Lat. Miles.

1 59.99 16 57.68 31 51.43 46 41.68|| 61 29.09 76 14.52 2 59.96 17 57.38 32 50.88 47 40.92 62 28.17 77 13.50 3 59.92 18 57.06 33 50.32 48 40.15 63 27.24 78 12.47 4 59.85 19 56.73 34 49.74 49 39.36 64 26.30 79 11.45 5 59.77 20 56.38 35 49.15 50 38.57 65 25.36 80 10.42 6 59.67 21 56.01 36 48.54 51 37.76 66 24.40 81 9.39 7 59.55 22 55.63 37 47.92 52 36.94 67 23.44 82 8.35 8 59.42 23 55.23 38 47.28 53 36.11 68 22.48 83 7.31 9 59.26 24 54.81 39 46.63 54 35.27 69 21.50 84 10 59.09 25 54.38 40 45.96 55 34.41 70 20.52|| 85 11 58.90 26 53.93 41 45.28 56 33.55 71 19.53 86 12 58.69 27 53.46 42 44.59 57 32.68 72 18.54 87 13 58.46 28 52.98 43 43.88 58 31.80 73 17.54 88 14 58.22 29 52.48 44 43.16 59 30.90 74 16.54 89 1.05 15 57.96 30 51.96 45 42.43 60 30.00 75 15.53 90 0.00

6.27 5.23 4.19

3.14

2.09

Let ABC represent a right-angled triangle; then, by Trig. onometry, Art. 41,

cos. B: R::AB: BC.

But, by the preceding Theorem, we have

:

Diff. Long.

A Departure

Lat.

B

cos. lat. R:: depart.: diff. long., from which we see that if one leg of a right-angled triangle represent the distance run on any parallel, and the adjacent acute angle be made equal to the degrees of latitude of that parallel, then the hypothenuse will represent the difference of longitude.

EXAMPLES.

1. A ship sails from Sandy Hook, latitude 40° 28' N., longitude 74° 1' W., 618 miles due east. Required her present longitude.

Cos. 40° 28': R:: 618: 812'.3=13° 32', the difference of Jongitude.

K

This, subtracted from 74° 1', leaves 60° 29′ W., the longi tude required.

2. A ship in latitude 40° sails due east through nine degrees of longitude. Required the distance run.

Ans. 413.66 miles.

of latitude 261 miles,

3. A ship having sailed on a parallel finds her difference of longitude 6° 15'. What is her latitude? Ans. Latitude 45° 54'.

4. Two ships in latitude 52° N., distant from each other 95 miles, sail directly south until their distance is 150 miles What latitude do they arrive at?

Ans. Latitude 13° 34'.

MIDDLE LATITUDE SAILING.

(199.) By the method just explained may be found the Iongitude which a ship makes while sailing on a parallel of latitude. When the course is oblique, the departure may be found by plane sailing, but a difficulty is found in converting this departure into difference of longitude.

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If a ship sail from A to B, the departure is equal to eb+fc +gd+hB, which is less than AC, but greater than DB. Navigators have assumed that the departure was equal to the distance between the meridians PA, PB, measured on a parallel EF, equidistant from A and B, called the middle latitude.

D

E

e

A

א

The middle latitude is equal to half the sum of the two extreme latitudes, if both are north or both south; but to half their difference, if one is north and the other south.

The principle assumed in middle latitude sailing is not perfectly correct. For long distances the error is considerable, but the method is rendered perfectly accurate by applying to the middle latitude a correction which is given in the accompanying tables, page 149.

(200.) It has been shown that when a ship sails upon an oblique course, the distance, departure, and difference of latitude may be represented by the sides of a right-angled trian

gle. The difference of longitude is derived from the departure, in the same manner as in parallel sailing, the ship being supposed to sail on the middle latitude parallel. Hence, if we combine the triangle ABC for plane sailing D with the triangle BCD for parallel sailing, we shall obtain a triangle ABD, by which all the cases of middle latitude sailing may be solved.

In the triangle BCD,

Cos. CBD : BC :: R: BD;

that is, cosine of middle latitude is to the departure, as radius is to the difference of longitude.

Diff. Long.

Lat

B

C

Departure

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In the triangle ABD, since the angle D is A

Distance

the complement of CBD, which represents the middle latitude, we have

Sin. D : AB :: sin. A : BD;

that is, cosine of middle latitude is to the distance, as the sine of the course is to the difference of longitude.

In the triangle ABC, we have the proportion
AC: BC:: R : tang. A.

But we have before had the proportion

Cos. CBD: BC :: R : BD.

The means being the same in these two proportions, we have Cos. CBD AC: tang. A: BD;

that is, cosine of middle latitude is to the difference of latitude, as the tangent of the course is to the difference of longitude.

The middle latitude should always be corrected according to the table on page 149. The given middle latitude is to be looked for either in the first or last vertical column, opposite to which, and under the given difference of latitude, is inserted the proper correction in minutes, which must be added to the middle latitude to obtain the latitude in which the meridian distance is exactly equal to the departure. Thus, if the middle latitude is 41°, and the difference of latitude 14°, the correction will be found to be 25', which, added to the middle latitude, gives the corrected middle latitude 41° 25'.

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