sailing these courses. Join CE, AC, AE; draw AB perpendicular to the meridian CF, and AD parallel thereto; then will AC=76.2 miles be the distance made good; AE 69.1 miles, the distance of Cape Cod from the ship; CE the dista ce of the two places= =136 miles; ACB=57° 36', the course made good; EAD=16° 34', the course to Cape Cod; and ECF the course from Mount-Desert rock to Cape Cod 38° 8', &c

EY LOGARITHMS.

To find the bearing and distance of the two places by Case V1. Plane Sailing.
To find the distance.

To find the bearing.
As difference of latitude 107... 2.02938
10.00000
Is to radius 45°
1.92428

As radius 90°..

So is departure 84.....

To tangent course 38° 8'...... 9.89490

10.00000 Is to difference of latitude 107. 2.02938 So is secant course 38° 8'...... 10.10426 To the distance 136

2.13364

......

distance

Whence the course from Mount-Desert rock to Cape Cod is S. 38° 8' W., 136 miles. The same may be found by the scale, or by inspection.

The difference of latitude and departure for the several courses being calculated, by Case 1. Plane Sailing, and arranged in the traverse table, it appears that the difference of latitude made good by the ship is 40.8 miles, and the departure 64.3 miles; then, by Case VI. Plane Sailing, these numbers are found to correspond to a course of S. 57° 36′ W. and distance 76.2 miles.

TRAVERSE TABLE.

Subtract the difference of latitude made good by the ship, 40.8 miles, from the whole difference of latitude, 107 miles, and there remain 66.2 miles, which is the difference of latitude between the ship and Cape Cod. In the same manner, by subtracting the ship's departure, 64.3 miles, from the whole departure, 84 miles, there remain 19.7 miles for the departure between the ship and Cape Cod. With this difference of latitude 66.2, and departure, 19.7, the bearing of Cape Cod is found, by Case VI. Plane Sailing S. 16° 34' W., and its distance, 69.1 miles.

All the preceding calculations may be made by logarithms, by the scale, or by mspection. But we shall leave them to exercise the learner, and for the same purpose shall add the following example.

EXAMPLE III.

A ship in the latitude of 37° 10′ N., is bound to a port in the latitude of 33° 0' N., which lies 180 miles west of the meridian of the ship; but by reason of contrary winds, she sails the following courses, viz. S. W. by W. 27 miles, W. S. W. W. 30 miles, W. by S. 25 miles, W. by N. 18 miles, S. S. E. 32 miles, S. S. E. 3 E. 27 miles, S. by E. 25 miles, S. 31 miles, and S. S. E. 39 miles. Required the latitude the ship is in, and her departure from the meridian, with the course and distance to her intended port.

The difference of latitude and departure made on each course, are given in the adjoined traverse table; hence it appears that the difference of latitude made good is 169.4 miles; the departure, 47.4 miles; and by Case VI. Plane Sailing, the course S. 15° 38′ W., and distance, 175.9 miles; and the course to the intended port, S. 58° 42′ W., distance 155.2 miles; the latitude being in 34° 21' N.

TRAVERSE TABLE.

*Instead of putting the course S. S. E. 32 miles, and S. S. E. 39 miles, you might make one entry

PARALLEL SAILING.

IN Plane Sailing, the earth is considered as an extended plane; but this supposition is very er oneous, because the earth is nearly of a spherical figure, in which the meridians all meet at the poles; consequently the distance of any two meridians measured on a parallel of latitude (which distance is called the meridian distance) decreases in proceeding from the equator to the poles. To illustrate this, let PB represent the semi-axis of the earth, B the centre, P the pole, PCA a quadrant of the meridian, AB the radius of the equator, and CD (parallel thereto) the radius of a parallel of latitude. Then it is evident that CD will be the cosine of AC, or the cosine of the latitude of the point C, to the radius AB; now, if the quadrantal arc PCA be supposed to revolve round the axis PB, the point A will describe the circumference of the equator, and C the circumference of a parallel of latitude; and the former circumference will be to the latter as AB to CD (as may easily be deduced from Art. 55, Geometry), that is, as radius to the cosine of the latitude, or the point C; hence it follows, that the length of any are of the equator intercepted between two meridians, is to the length of a corresponding are of any parallel intercepted between the same meridians, as radius is to the cosine of the latitude of that parallel. Hence we obtain the following theorems.

THEOREM I.

A

B

The circumference of the equator is to the circumference of any other parallel of latitude, as radius is to the cosine of that latitude.

THEOREM II.

As the length of a degree of the equator is to the meridian distance corresponding to a degree on any other parallel of latitude, so is radius to the cosine of that parallel oʻ

Latitude.

THEOREM III.

As radius is to the cosine of any latitude, so are the miles of difference of longitude between two meridians (or their distance in miles upon the equator) to the distance of these two meridians on that parallel of latitude in miles.

THEOREM IV.

As the cosine of any latitude is to radius, so is the length of any arc on that parallel of latitude (intercepted between two meridians) in miles to the length of a similar arc on the equator, or miles of difference of longitude.

THEOREM V.

As the cosine of any latitude is to the cosine of any other latitude, so is the length of any are on the first parallel of latitude in miles, to the length of the same arc on the other

in miles.

By means of Theorem III. the following table was calculated, which shows the meridian distance corresponding to a degree of longitude in every latitude; and may be made to answer for any degree or minute by taking proportional parts.

The following Table shows for every degree of latitude how many miles distanı the two meridians are, whose difference of longitude is one degree.

LAT

MILES. LAT. MILES. LAT. MILES. LAT. MILES. LAT. MILES.

When a ship sails east or west on the surface of the earth supposed to be spherical, she describes a parallel of latitude, and this is called Parallel Sailing. In this case, the distance sailed (or departure) is equal to the distance between the meridians sailed from and arrived at in that parallel; and it is easy, by Theorem IV. (preceding) to find the difference of longitude from the distance, or the distance from the difference of longitude, as will appear plain by the following examples.

CASE I.

The difference of longitude between two places in the same parallel of latitude being given, to find the distance between them.

Suppose a ship in the latitude of 49° 30′, north or south, sails directly eas !r west, until her difference of longitude be 3° 30′; required the distance sailed.

BY PROJECTION.

Take the sine of 90° from the plane scale, and, with one foot of the compasses on (fig. 1) as a centre, describe the arc EQ with the difference of longitude, 210 miles. in the compasses, and one foot in E,

as a centre, describe an arc cutting EQ in Q; jom PE, PQ. Take the sine of the complement of the latitude 40° 30′ in your compasses, and with one foot in P, as a centre, describe the arc FG, cutting PE, PQ, in F, G; then the length of the chord FG being measured on the same scale of equal parts, will be the departure 136.4 miles.

Or this projection may be made in the following manner. Draw AD (fig. 2) of an indefinite length; make the angle DAC equal to the latitude 49° 30', and AC equal to the difference

E

F

P

Diff Long. 210.

Sine 90°
FIG. 1.

of longitude 210 miles; draw CD perpendicular to AD; then will the line AD be the distance or departure required.

BY LOGARITHMS.
To find the departure or distance.

As radius 90°
Is to the difference of longitude 210.
So is cosine latitude 49° 30′.......

10.00000

2.32222

9.81254

BY GUNTER

The extent from radius to the complement of the latitude 40° 30' on the line of sines, will reach from the difference of longitude 210, to the distance 136.4, on the line of numbers.

BY INSPECTION.

Find the latitude among the degrees in Table II., and in the distance column the difference of longitude, opposite to which in the column of latitude will be the distance required.

In the present example, the latitude is 49° 30′; and as the table is only calculated to single degrees, we must find the numbers in the tables of 49° and 50°, and take the mean of them; the former is 137 8, the latter 135.0, the mean of which is the sought distance or departure, 136.4.

CASE II.

The distance between two places on the same parallel of latitude given, to find the difference of longitude.

Suppose a ship in the latitude of 49° 30′ N. or S., and longitude 36° 40′ W., directly west 136.4 miles; required the difference of longitude, and longitude in.

BY PROJECTION.

sails

With the sine of the complement of the latitude, 40° 30', in your compasses, and one oot in P, as a centre (fig. 1, of the preceding case), describe the arc FG, upon which set off the departure 136.4 miles, upon the chord FG, and through the points F and G Iraw the lines PE and PQ; then, with the sine of 90° in the compasses, and one foot n P, as a centre, describe an arc to cut PE, PQ, in E and Q; then the chord EQ eing measured upon the same scale of equal parts that the departure was, will be the ifference of longitude 210 miles.

Or thus; draw the line AD (fig. 2), which make equal to the given distance 136.4; D erect DC perpendicular to DA; make the angle DAC equal to the latitude; then will AC be the sought difference of longitude 210 miles.

Look for the latitude among the degrees, as if it was a course, and the departure in the column of latitude; against which will stand the difference of longitude in the distance column.

Thus, in the course 49°, we must seek for 136.4 in the latitude column, and we find it corresponds to the distance 208; and in the course 50°, we find it nearly corresponds to 212; half the sum of 208 and 212 is 210, which is the sought difference of longitude.

QUESTIONS

To exercise the learner.

Question I. A ship in the latitude of 32° N., sails due east till her difference of longitude is 384 miles; required the distance sailed.

Answer. 325.7 miles.

Quest. II. A ship from the latitude of 53° 36′ S., longitude 10° 18′ E., sails due west 236 miles; required her present longitude.

Ans. 3° 40' E.

Quest. III. If two ships in the latitude of 44° 30′ N., distant 216 miles, should sail directly south until they were in the latitude of 32° 17′ N., what distance are they from each other?

Ans. By Theorem V., 256 miles.

Quest. IV. A ship having run due east for three days, at the rate of 5 knots an hour, finds she has altered her longitude 8° 16'; what parallel of latitude did sh sail in:

Ans. 43° 28′ N. or S.

MIDDLE LATITUDE SAILING.

In sailing north or south (or on a meridian) the difference of longitude is nothing, and the difference of latitude is equal to the distance sailed; but in s ding east or wes for on a parallel of latitude), the difference of latitude is nothing, and the difference of longitude may be calculated by the foregoing theorems of Parallel Sailing. In sailing on any other course, the ship changes both her latitude and longitude; in this case the duference of latitude, departure, and difference of longitude, may be calculated by a proper application of the principles of Plane Sailing to the sailing on a spherical surface; to do which, the surface of the globe may be supposed to be divided into an indefinite number of small surfaces, as square miles, furlongs, yards, &c., which, or account of their smallness, in comparison with the whole surface of the earth, may be esteemed as plane surfaces, and the difference of latitude and departure (or meridian distance) made in sailing over each of these surfaces, may be calculated by the common rules of Plane Sailing; and by summing up all the differences of latitude and departures made on these different planes, we shall obtain the whole difference of latitude and departure nearly.* Now, by Case I. of Plane Sailing, the distance described on any one of these small surfaces is to the corresponding difference of latitude as radius is to the cosine of the course; and as the course is the same on all these surfaces, it follows that the sum of all the distances described thereon, is to the sum of the corresponding differences of latitude as radius is to the cosine of the course; that is, the whole distance sailed on the globe, is to the corresponding difference of latitude as radius is to the cosine of the course. In a similar manner it appears, that the distance described on the globe is to the sum of all the corresponding departures (or meridian distances) described on these different surfaces, as radius is to the sine of the course; so that the canons for calculating the whole difference of latitude and departure from the course and distance are the same, whether the earth be esteemed as an extended plane or a spherical surface; and the same is to be observed with respect to the other cases of Plane Sailing.

We shall, therefore, in all the calculations of sailing on the spherical surface of the earth, in which the course, distance, difference of latitude and departure, occur, make use of the canons already taught in Plane Sailing, and shall construct the schemes exactly in the same manner. The only additional calculation in sailing on a spherica! surface, consists in determining the longitude from the departure; for in sailing on a plane, the departure and longitude are the same; but in sailing on a spherical surface, the whole departure (as was observed above) is equal to the sum of all the meridian distances made in sailing over the indefinite number of small surfaces, into which we have supposea the spherical surface to be divided, and the whole difference of longitude corresponding is equal to the sum of all the differences of longitude, deduced from each of these small meridian distances by Theorem IV. of Parallel Sailing. Several methods have been proposed for abridging the calculation of the difference of longitude from the departure, the most noted of which are those known by the names of Middle Latitude Sailing and Mercator's Sailing; the latter (which will be hereafter explained) is perfectly accurate; the former is only an approximation, but it is very much used in calculating

*The error arising from this supposition will be decreased by increasing the number of the planes, so that, by increasing the number indefinitely, the error may be made less than any assignable quantity, Using (in estimating the difference of longitude corresponding to each of these small meridian distances) the latitude corresponding to the middle point of the surface on which these small meridian distances are respectively made.

This is true in theory, and would be so in practice, if the meridional difference of latitude i Table III. were given to a sufficient number of decimals; but being only given to the nearest mile or minute, the error arising from this cause, when the difference of latitude is small, is greater than the error in Middle Latitude Sailing; in consequence of this, the method by middle latitude is almos