QUESTIONS FOR EXERCISE.

Question 1. A ship in the latitude of 49° 57′ N., and longitud: of 15 – 16′ W., £ails south-westerly until her departure is 789 miles, and then, by observation, is in the latitude of 39° 20 N.; required her course, distance, and longitude in.

Answer. Course S. 51° 05′ W., distance 1014 miles, longitude in 33° 50′ W.

Quest. II

A ship in the latitude of 42° 30′ N., and longitude of 58° 51′ W., sais S. W. by S. 591 miles; the latitude, and longitude in, are required.

Ans. Latitude in 34° 19′ N., longitude in 65° 51′ W.

Quest. III. A ship from the latitude of 49° 57' N., and longitude of 30° 00′ W., sails S. 39 W. till she arrives in the latitude of 45° 31′ N.; required the distance run, and longitude in.

Ans. Distance 342.3, longitude in 35° 21′ W.

Quest. IV. A ship from the latitude of 50° 10′ S., and longitude of 30° 00′ E., sails E. S. E. until her departure is 957 miles; required the distance sailed, and the latitudo and longitude in.

Ans. Distance 1036 miles, latitude in 56° 46′ S., longitude in 56° 50′ E.

Quest. V. A ship in the latitude of 49° 30′ N., and the longitude of 25° 00′ W., sails south-easterly 645 miles, making 500 miles departure; required the course steered, and the latitude and longitude in.

Ans

Course S. 50° 49′ E., latitude in 42° 42′ N., longitude m 12° 57′ W.

Having gone through the necessary problems in Mercator's Sailing, we shall now show how Mercator's Chart may be constructed by means of the Table of Meridional Parts.

To construct a Mercator's Chart to commence at the equator.

Suppose it was required to construct the Chart in the Plate prefixed to this work, which begins at the equator, and reaches to the parallel of 50 degrees, and contains 95 degrees of longitude west from the meridian of Greenwich.

Draw the line AD representing the equator; then take from any scale of equal parts the number of minutes contained in 95 degrees, viz. 5700, which set off from A to D; subdivide this line into 95 equal parts, representing degrees of longitude. Through A and D draw the lines AB, DC, perpendicular to AD, and make each of them equal to 3474, which are the meridional parts, corresponding to 50 degrees. Join BC, which must be subdivided in the same manner as the line AD; and through the corresponding points of the lines AD, BC, must be drawn (at the distance of 10° or 20°) the lines parallel to AB, representing meridians of the earth; these lines must be numbered 0, 10, 20, &c., beginning at the line AB, which represents the meridian of Greenwich. Set off in like manner upon the meridians AB, DC (beginning from the equator AD), the meridional parts corresponding to each degree of latitude from 0° to 50°; and through the corresponding points (at the distance of 10° or 20°) draw lines parallel to the equator AD, to represent the parallels of latitude. Then the upper part of the chart will represent the north, the lower the south, the right hand the east, and the left hand the west (which is generally supposed in charts, unless the contrary is expressly mentioned).

If the chart does not commence at the equator, but is to serve for a certain portion of the globe contained between two parallels of latitude on the same side of the equator, you must draw the meridians as directed in the last example; then subtract the meridional parts of the least latitude of the chart from the meridional parts of the other latitudes, and set off these differences on the extreme meridians; draw lines through the corresponding points, and they will be the parallels of latitude on the chart.

If the chart is to be bounded by parallels of latitude on different sides of the equator, you must draw a line representing the equator, and perpendicular to it draw the lines to represent the meridians, continuing them on both sides of the equator; then set off the prallels of latitude on both sides of the equator, in the same manner as in the first example.

Take from the Table of Latitudes and Longitudes of places the latitude and longitude of each particular place contained within the bounds of the chart, and lay a rule over its latitude, and another crossing that over its longitude; the point where these meet will represent the proposed place upon the chart. The most remarkable point of a seacoast being thus laid down, lines may be drawn from point to point, which will form the outlines of the sea-coast, islands, &c.; to which may be annexed the depths of water expressed in common Arabian numbers, the time of high water on the full and change days expressed in Roman numbers, the setting of the tide expressed by an arrow, and whatever else may be thought convenient for the chart to contain.

This chart is not to be considered as a just representation of the earth's surface, for the figures of islands and countries are distorted towards the poles, as is evident from the construction; but the degrees of latitude and longitude are increased in the same proportion, so that the bearings between places will be the same on the chart as on the globe; and as the meridians are right lines, it follows, that the rhumos, which form equal angles with the meridians, will be straight lines, which render this projection of the earth's surface much more easy and proper for the mariner's use than any other.

Having the latitude and longitude of a ship or place, to find the corresponding point on the chart.

RULE. Lay a ruler across the chart in the given parallel of latitude; take in your compasses the nearest distance between the given longitude and the nearest meridian drawn across the chart; put one foot of the compasses in the point of intersection of the ruler and meridian, and extend the other along the edge of the ruler on the same side of the meridian as the place lies, and that point will represent the place of the ship. If the longitude on the chart be counted from a different meridian from that you reckon from, you must reduce the given longitude to the longitude of the chart, by adding or subtracting the difference of longitude of those meridians, and then mark off the ship's place, as before directed. Or you may draw a meridian line through the place you reckon your longitude from; then measure off the ship's longitude on the equator, and apply it to the edge of the ruler from this meridian, and you will obtain the ship's place.

To find the bearing of any place from the ship.

RULE. Lay a ruler across the given place and the place of the ship; set one foot of the compasses in the centre of some compass near the ruler, and take the nearest distance to the edge of the ruler; slide one foot of the compasses along that edge, keeping the other extended to the greatest distance from the ruler, and observe what point of the compass it comes nearest to, for that will be the bearing required.

To find the distance of any place from the ship.

RULE. Take the distance between the ship and the given place in your compasses, and apply it to the side of the chart or graduated meridian, setting one foot as much above one place as the other is below the other place; the number of degrees between the points of the compasses will be the distance nearly.

When the places bear north and south of each other, this rule is accurate; but when they bear nearly east and west, and the distance is large, it will err considerably; but in general it is exact enough for common purposes; if greater accuracy is required, it is best to find the distance by calculation.

If any one wishes to estimate the distance accurately by the chart, he must proceed in the following manner:

1. If the place be in the same longitude that the ship is in, then the preceding rule

is accurate.

2. If the place be in the same latitude as the ship, or bear east or west, the distance cannot be obtained without calculating it by Case I. of Parallel Sailing.

3. If the place be neither in the same latitude, nor in the same longitude as the ship. the distance must be found in the following manner:-Lay a ruler over both places, and draw through one of them a parallel to the equator; take the difference of latitude between both places in your compasses from the equator; slide one foot on that parallel, keeping the other extended so that both points shall be on the same meridian, and note the point of the ruler which is touched by the other foot of the compasses; take the distance from this point to the given place through which the parallel was drawn, and apply it to the equator, and you will have the sought distance.

The bearing and distance of any two places from each other may be found in the same manner as the bearing and distance of any place from the ship.

EXAMPLE.

Required the bearing and distance between the east end of Long Island and the north part of Bermudas.

A ruler being laid over both places, as directed in the preceding rule, it will be found to lie parallel to the N. W. by N. and S. E. by S. line; and the distance between the two places being taken in the compasses, and applied to the graduated meridian, will measure about 10 degrees or 600 miles; therefore these places bear from each

PROBLEMS USEFUL IN NAVIGATION AND

SURVEYING.

PROBLEM I.

Loasting along shore, I saw a cape of land bearing N. N. E., and after sailing W. N. W 20 miles, it bore N. E. by E.; required the distance of the ship from the cape at both

stations.

BY PROJECTION.

Describe the compass ESW, and let its centre A represent the place of the ship at the first station; draw the W. N. W. line AB equal to 20 miles, and B will represent the second station. Draw the N. N. E. line AC, of an indefinite length, and the line BC parallel to the N. E. by E. line of the compass; the point of intersection C will represent the place of the cape; and the distance BC, being measured, will be found 36 miles; and AC 30 miles.

BY LOGARITHMS, (BY CASE I. OF OBLIQUE TRIGONOMETRY.) The difference between N. N. E. and W. N. W. is 8 points or 90°, therefore BAC is a right angle; also the difference between the N. E. by E. and N. N. E. is 3 points, equal to the angle ACB; and the difference between the N. E. by E. point and the point opposite to W. N. W. is 5 points, equal to the angle ABC.

To find the distance BC.

......

As sine angle ACB 3 pts. Ar. Co. 0.25526 Is to the distance AB 20 1.30103 So is sine angle BAC 8 points.. 10.00000 To the distance BC 36.0 1.55629

......

To find the distance AC.

As sine ACB 3 points...Ar. Co. 0.25526
Is to the distance AB 20....... 1.30103
So is sine angle ABC 5 points .. 9.91985
To the distance AC 29.93...... 1.47614

The above solutions are by Case I. Oblique Trigonometry, though they might have been done, in this example, by Case II. of Right-angled Trigonometry, because the angle BAC is a right angle.

G H

If the bearings of the middle point C of an island (or any remarkable peak) be determined in this manner, we may, at the same time, find the limit of the dimensions of the island, by measuring with a quadrant or sextant, held in a horizontal position, the angular distances between that middle point and the extremes of the island. For by drawing the lines ADE, AGF, making the angles DAČ, GAC, with AC, equal to the angular distances observed at A. and in the same manner by drawing the lines BDG, BEF, making angles with BC equal to the angular distances observed at B, you would obtain the quadrilateral figure DEFG, within which the island is to be placed. If similar observations could be procured at H, they would in general take off the corners at D and F; and observations at I would generally take off the corners at E and G; and by observing the projecting points and coves in the island, while sailing round it, and drawing a figure conformable thereto, within the limiting space thus found, the form and dimensions of the island may be obtained to a considerable degree of accuracy.

D

B

I

PROBLEM II.

being at sea, we saw two headlands, whose bearing from one another by the chart was W. by N., and E. by S., and the distance 15 miles; the westernmost bore from us S. S. W., and the easternmost S. E. by E.: required our distance from each of them.

BY PROJECTION.

N

Draw the compass NESW, and through the centre A, draw the E. by S. line AR, the S. S. W. line AB, and the S. E. by E. line AC, and continue the two latter indefinitely; upon the former, AR, take AD equal to 15 miles; through D draw DC parallel to AB, to meet AC in C, and draw CB parallel to AD. Then A will be the place where the headlands B and C were observed; and the distance AB of the westernmost headland, being measured, is found to be 5.8 miles, and the distance AC of the easternmost headland 15 miles.

W

BY LOGARITHMS.

E

Ꭼ Ꮟ Ꮪ

DR

SEVE

B

Ebs 15

S

Between the S. S. W. line AB, and the S. E. by E. line AC, are 7 points = angle BAC; and between the S. E. by É. line AC, and the E. by S. line AD, are 2 points= angle CAD angle ACB (because AD, BC, are parallel); therefore ACB+BAC=9 points; and since all three angles ACB, BAC, ABC, are equal to 16 points, the angle ABC is also equal to 7 points; therefore (by Art. 39, Geometry) the sides AC, CB, are equal, being opposite to the equal angles ABC, BAC. If these angles had not been equal, the side AC might have been calculated in the same manner as we shall now calculate the side AB.

To find the side AB.

This problem, or the first, may be used for finding the distance of a ship from any headland, &c., when taking a departure from the land.

PROBLEM III.

Two ships sail from the same port; the first sails N. E.

E. 16 miles; the second sails

easterly 20 miles, and then finds that the first bears N. N. W.: required the course of the second ship, and the distance between the two ships.

BY PROJECTION.

Draw the compass ESW, and let its centre A represent the port sailed from draw the N. E. E. line AB equal to 16 miles; also through B, the line BC, parallel to the N. N. W. line, and continue it indefinitely; take a distance representing 20 miles in your compasses, and putting one foot in A, describe with the other an arc cutting the line BC in C, and join AC. Then B will be the place of the first ship, C that of the second, and AC the course steered by the second ship, which will be nearly E. S. E. E., and BC the distance of the ships 17 miles.

The course from B to C is S. S. E. (opposite to N. N. W.), and from B to A 18 S. W. W. (opposite to N. E. E.); the difference between these bearings is 64 points, equal to 73° 7', equal to the angle ABC; having this angle and the sides AB, AC, we may find the other angles and. side by Cases II. and III. of Oblique

Two ships sail from the same port, the one N. W. 30 miles, and the other N. E. by N. 40 miles; required the bearing and distance of the ships from each other.

BY PROJECTION.

Draw the compass NESW, and let its centre A represent the port sailed from; draw the N. W. line AB equal to 30 miles, and the N. E. by N. line AC equal to 40 miles; join BC, which will be the bearing and distance of the two ships; whence the bearing will be found to be W. S. W. § W., and the distance 45.1 miles, nearly.

BY LOGARITHMS, (BY CASES IV. V. OF OBLIQUE TRIGONOMETRY.) Between the N. W. line AB, and the N. E. by N. line AC, there are 7 points, equal to angle BAC; half the supplement of this to 180° is 50° 37', equal to half sum of the angles C and B.

To the angle C, equal to 40° 45', add the angle representing the course from C to A, equal to 33° 45', the sum is 74° 30′, which is the bearing of B from C, namely, S. 74° 30′ W., or W. S. W. W., nearly.

PROBLEM V.

Two ports bear from each other E. by N. and W. by S., distance 400 miles: a ship from the easternmost sails northerly 450.7 miles; another from the westernmost sails 300 miles, and meets the first: required the course steered by each ship.

BY PROJECTION.

Draw the compass ESW, and let the centre B represent the westernmost port; draw the E. by N. line BD equal to 400 miles, and D will be the easternmost port; with 300 in your compasses, and one foot in B, describe an arc; with 450.7 in your compasses, and one foot in D, describe another arc, cutting the former in C; join DC, BC. Then BC will be the course sailed by the westernmost ship, and DC the course sailed by the easternmost ship.