The ship is in lat. 34° 21' N. the dep. is 47.4 W.
The cou. made good is S. 15° 38' W. and dist. 175.9.

The cou to the intended port is S. 58° 43′ W. or S. W. by W. one quarter west nearly, distance 155.2.

MIDDLE LATITUDE SAILING.

IN Plane Sailing the earth was considered as a plane, representing a bowling-green, having the meridians parallel to each other, aud consequently the degrees of longitude equal in all places; but this. cannot be true, as the earth is a globe or sphere; for,

As the meridians are circles on the terraqueous globe, meeting in the poles (as may be seen in the Plate page 46), it is obvious, that any two of those circles must recede more at greater distances from the poles; and at equal distances from each pole, or at the equator, the distance between the meridians is greatest.

The true place of a ship at sea depends upon its distance from the equator, and some noted meridian; and since the meridional distance, that is, the distance between any two meridians, varies in every latitude, it is therefore convenient this distance should be reckoned in a fixed latitude, and where the degrees are of the saine magnitude with those of the meridian, which can be no where but on the equator, where 60 geographical miles make a degree.

The circumferences of all circles are in direct proportion to each other, as their radii; and since the earth turns once round its axis in 24 hours, every point upon its surface must describe circles parallel to the equator: hence it follows, that the circumference of any parallel of latitude, in miles, is to the circumference of the equator, in miles, as the co-sine of that latitude is to radius; and, that the breadth of a degree, in any parallel of latitude, is to the breadth of a degree upon the equator, as the sine complement of that latitude is to radius.

By the last proportion was the following Table calculated, which shows the breadth of a degree of longitude in every latitude; and may be made to answer for any degrees or minutes by taking proportional parts,

The following Table shows how many Miles answer to a Degree of Longitude at every Degree of Latitude.

D. L MILES. D. L. MILES. D. L. MILES. D. L. MILES. D. L. MILES.

15..52|

14..51

13

.50

12..48

1 59.99 19 56..73 37 47..92 55 34..41 73 17..5+ 259.96 20 56..38 38 47..28 56 33.55 74 16..53 3 59.92 21 56..01 39 46..62 57 32..68 75 459..86 22 55..6 40 45..95 58 31..79 76 559.77 23 55..23 41 45..28 59 30..90 77 6 59.67 24 54..81 42 44..59 60 30..00 78 759..56 25 54.38 43 43..88 61 29.09] 79 [11..45] 859.42 26 53.93 44 43..16 62 28..17 80 10..42) 9 59.26 27 53.46 45 42. .43 63 27..24 81 10 59.08 28 52..97 46 41..68 64 26..30 82 1158.89 29 52..47 47 40..92) 65 (25..36 83 12 58.68 30 51..96 48 40..15 66 24.4 84 13 58.46 SI 1..43 49 39..36 67 23.45 85 14 58..22 32 50..88 50 38..57 68 22. .48 15 37..95 33 50.32) 51 37..76 69 21.50 87 16 57..67 34 49..74 52 36.94 70 20..52 88 17 57..37 35 49..15 53 36..11 71 19.54 89 18 (57..06 36 48..54 54 35. .26 72 18.55)

Hence it follows, that

As radius, or sine 90°
Is to the diff. of long. in miles,
So is co-sine of any paral. of lat.
To the dist. in miles between any
two mer. in that paral. of lat,

From what has been said, arises
Problems,

9..38

8..35

7..32

6..28

86

5..26 4.18

3..14

2..09 1..05

As co-sine of any paral. of lat.
Is to the distance run in miles
in that lat.

So is the radius, or sine of 90o
To the diff, of long. in miles.

the solution of the following

PROBLEM I.

The Difference of Longitude between two Places, both in one Paral lel of Latitude, being given, to find the Distance between them. Suppose a ship in the lat. 49° 30' N. or S. sails directly E. or W. until her diff, of long. be 3o 30', and the dist, sailed be required.

With the sine of 90° in your compasses, taken from the Plane Scale, and with one foot in P, describe the arch EQ, and upon it set off the diff. of long. 210 miles, and draw the lines PE and PQ to represent the two meridians; and then EQ represents the equator, and P the pole. Again, with the sine com. of the lat. 49° 30, viz. 40° 30' in your compasses, taken from the line of sines on the Plane Scale, and with one foot in P, describe an arch, and the dist. between the points, where it cuts the two meridians, being measured upon the same scale of equal parts that the diff, of long, was, will be the dep. 136.4 miles.

Or, thus:

Draw the mer. AB, and with the chord of 60° in your compasses describe an arch, and upon it set off the comp. of the lat. 40° 30' (taken from the line of chords), and set it off upon the arch as a cou. in Plane Sailing, and draw the line AC as a dist. which make equal to the diff. of long. 210 miles; then will the departure CD be the distance 136.4 miles as before: this last method is preferable to the former, as we are not confined to any particular scale.

Reverse this Problem, and suppose the dist. sailed in any parallel of lat. given, to find the diff. of long.

With the sine com. of lat, in your compasses describe an arch, upon which set off the dep. 136.4 miles, and through the points where it cuts the arch draw the lines PE and PQ; then, with the sine of 90° in your compasses, and one foot in the former centre'P, describe an arch to cut PE and PQ; then EQ being measured upon the small scale of equal parts that the dep. was, will be the diff. of long, 210 miles.

The extent from rad. to sine com. lat. 40° 30' on the line of sines will reach from the diff. of long. 210 to the dist. 136.4 on the line of numbers.'

BY INSPECTION.

Find the sine com. of the lat. among the degrees, and in the dist. column the diff. of long., opposite to which, in the column of dep. is the dist. required; but as the co-lat. is 40° 30', therefore,

For 40 degrees you will find
For 41 degrees you will find

The sum is

Half the dist. required

135

137.8

272.8

136.4

This is done because the Table of Diff. of Lat. and Dep. is calculated only for single degrees.

By the reverse of the last problem, having the dist. run in any parallel to find the diff. of long.

Suppose a ship in lat. 49° 30' N. or S. sails directly E. or W. 136.4 miles, and her diff. of long. be required.

Look for the comp. of the lat. among the degs. as if it was a cou. and the dep. in its column: right against which stands the diff. of long, in the dist. column. In the last Problem the ship is supposed to have sailed due east or west, in the same parallel of lat.; "but in her course she generally crosses several meridians and parallels,and then arrives at a different lat. from that she left; and, as it is plain

by the foregoing Table, that the miles which make a degree in one parallel, will not be the same as those that make a degree in any other parallel, lying on the same side of the equator; therefore add both lats. together, and take half their sum for a mean or mid. lat.; which may be conceived as if the ship had sailed in one lat.; with which the diff. of long. may be turned into dep. and dep. into diff. of long. in the same manner as has been already shown, for it will be

Having the diff. of lat. and dep., the cou. and dist. are found by Case the Sixth, in Plane Sailing.

CASE I.

Required the bearing and dist. between the Lizard, in lat. 49° 57' N. long. 5° 12' W. and the island of St. Mary, one of the Western islands, in lat. 36o 58' N. and long. 25° 12' W.

Draw the mer. AE, with the chord of 60 describe the arch PS; upon which set off 46° 32', the comp. of mid. lat. from Q to S; through S draw the line AC-1200, the diff. of long. let fall the perpendicular CE, which will be the dep. 870.9; upon AE set off AD 779, the diff. of lat. ; and upon Derect the perp. DG, and