To find the Middle Latitude = the Angle BCD:-

With the departure = BC, and the difference of longitude = CD, the angle of the middle latitude may be found by trigonometry, Problem III., page 174; as thus:

With the angle of the course, thus found, and the difference of latitude AB, the distance may be computed by trigonometry, Problem IV., pàge

Hence, the course is S. 46:54. W., or S.W. W. nearly, and the distance 747.8 miles.

To find the Latitude come to, Course, and Distance, by Inspection in the general Traverse Table :

=

One-fourth of the difference of longitude = 2234, taken as distance, and one-fourth of the departure 136. 5, in a latitude column, will be found to agree between 52 and 53:. Now, to latitude 52°, and distance 223, the difference of latitude is 137.3, which is 0'. 8 more than 136.5; and to latitude 53%, and distance 223, the difference of latitude is 134. 2, being 2'.3 less than 136.5: hence, the difference of meridional distance to 1: of latitude is 0'. 8+2'.33'.1: therefore, as 3'.10'.8 :: 60:: 16., which, added to 52: (proportion being made for the quarter of a mile in the distance), gives the middle latitude = 52:18: hence, the latitude come to is 56:34: S., and the difference of latitude 511 miles. Again, to one-fourth of the difference of latitude 127.75, and one-fourth of the departure 136. 5, the course is 47%, and the distance 187; which, multiplied by 4, gives 748 miles the whole distance.

PROBLEM IX.

Given the Distance, Difference of Longitude, and Middle Latitude; to find the Course and both Latitudes.

Example.

A ship, in north latitude, sailed 500 miles upon a direct course between the south and west, until her difference of longitude was 440 miles; required the course steered, the latitude sailed from, and the latitude come to; allowing the middle latitude to be 43:45: north?

To find the Angle of the Course = A:

The course may be found by the 3d analogy, page 222, as thus:

To find the Difference of Latitude A B:

The difference of latitude may be found by the 8th analogy, page 222, as

SOLUTION OF PROBLEMS IN MERCATOR'S SAILING.

Mercator's Sailing is the method of finding, on a plane surface, the motion of a ship upon any assigned point of the compass, which shall be true in latitude, longitude, and distance sailed.

Mariners, generally speaking, solve all the practical cases in Mercator's Sailing by stated rules, called canons, which they early commit to memory, and, ever after, employ in the determination of a ship's place at sea. Those canons, certainly, hold good in most cases; but since they are destructive of the best principles of science, inasmuch as that they have a direct tendency to remove from the mind every trace of the elements of trigonometry, the very doctrine from which they were originally deduced, and on which the whole art of navigation is founded, the following observations and consequent analogies are, therefore, submitted to the attention of naval people, under the hope that they will serve as an inducement to the substitution of the rules of reason for the rules of rote; and thus do away with the necessity of getting canons by heart.

In the annexed diagram, let the triangle ABC be a figure in plane sailing, in which the angle A represents the course, A C the distance, AB the difference of latitude, and B C the departure. If A B be produced to D, until it is made equal to the meridional difference of latitude, and D E be drawn at right angles thereto, and parallel to BC; then the triangle ADE will be a figure in Mercator's sailing, in which the angle A represents the course, the side AD the meridional difference of E latitude, and the side D E the difference of longi

C

Depart

Diff long

A

Course

Diff lat.

Mer diff lat

tudę. Now, since the two triangles ABC and ADE are right angled,

B

D

and that the angle A is common to both; therefore they are equiangular and because they are equi-angular, they are also similar; therefore the sides containing the equal angles of the one are proportional to the sides containing the equal angles of the other.-Euclid, Book VI., Prop. 4.

Now, from the relative properties of those two triangles, all the analogies for the solution of the different cases in Mercator's sailing may be readily deduced agreeably to the established principles of right angled trigonometry, as given in page 171, and thence to 177; as thus :

First, in the triangle ABC, if the distance AC be made radius, the analogies will be,

1. As radius distance AC :: sine of the course A: departure BC; and co-sine of the course A: difference of latitude A B.

2. As sine of the course A: departure BC: radius distance AC; and co-sine of the course A difference of latitude A B.

3. As co-sine of the course A difference of latitude AB radius: distance A C; and :: sine of the course A: departure BC.

4. As the distance AC radius :: departure BC: sine of the course A; and difference of latitude AB: co-sine of the course A.

Again, by making the difference of latitude A B radius, the analogies will be,

5. As the difference of latitude AB: radius: departure BC: tangent of the course A; and :: distance AC: secant of the course A.

6. As radius difference of latitude AB:: tangent of the course A: departure BC; and secant of the course A: distance A C.

And by making the departure B C radius, it will be,

7. As the departure BC radius: difference of latitude AB co-tangent of the course A; and distance AC: co-secant of the course A.

8. As radius departure BC: co-tangent of the course A: difference of latitude A B; and co-secant of the course A: distance A C.

Now, in the triangle A D E, if the meridional difference of latitude A D be made radius, the analogies will be,

9. As the meridional difference of latitude AD: radius :: difference of longitude D E tangent of the course A.

10. As radius meridional difference of latitude AD: tangent of the course A difference of longitude D E.

And by making the difference of longitude D E radius, it will be,

11. As the difference of longitude DE: radius :: meridional difference of latitude D E co-tangent of the course A.

12. As radius difference of longitude DE:: co-tangent of the course A meridional difference of latitude A D.

Finally, since the triangles A B C and ADE are equi-angular and similar, we have,

13. As the difference of latitude AB departure BC: meridional difference of latitude AD: difference of longitude D E.

The meridional difference of latitude is found by means of Table XLIII., by the same rules as those for the difference of latitude given at page 214; as thus :-If the two given latitudes be of the same name, the difference of their corresponding meridional parts will be the meridional difference of latitude; but if the latitudes be of contrary names, the sum of these parts will be the meridional difference of latitude.

PROBLEM I.

Given the Latitudes and Longitudes of two Places; to find the Course and Distance between them.

To find the Course = Angle A:

This comes under the 9th analogy, in page 237: hence,

= 126 miles.