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the parallels of latitude, and at right angles to these, draw. the meridians in such a manner, that their distance from each other shall be to the distance of the parallels of latitude, as the cosine of the latitude of the middle of the chart, to radius.

After the lines on all the sides are graduated, the positions of the several places which are to be laid down, may be determined, by applying the edge of a rule, or strip of paper, to the divisions for the given degree of longitude on each side, and another to the divisions for the degree of latitude. In the intersection of these, will be the point required.

The distance which a ship must sail, in going from one place to another, on a single course, may be nearly found, by applying the measure of the interval between the two places, to the scale of miles of latitude on the side of the chart.*


83. In Mercator's chart, the meridians are drawn at equal distances, and the parallels of latitude at unequal distances, proportioned to the meridional differences of latitude. (Arts. 63, 67.) To construct this chart, then, make a scale of equal parts on one side of the paper, for the lowest parallel of Ĩatitude which is to be laid down, and divide it into degrees and minutes. Perpendicular to this, and through the dividing points for degrees, draw the lines of longitude. For the second proposed parallel of latitude, find from the table, (Art. 67.) the meridional difference of latitude between that and the parallel first laid down, and take this number of minutes from the scale on the chart, for the interval between the two parallels. In the same manner, find the interval between the second and third parallels, between the third and fourth, &c. till the projection is carried to a sufficient extent.

Places whose latitudes and longitudes are known, may be laid down in the same manner as on the plane chart, by the intersections of the meridians and lines of latitude passing through them.

If the chart is upon a small scale, the least divisions on the graduated-lines may be degrees instead of minutes; and the meridians and parallels may be drawn for every fifth or every tenth degree. But in this case it will be necessary to

* See Note G.

divide the meridional differences of latitude by 60, to reduce them from minutes to degrees.

84. The Line of Meridional Parts on Gunter's scale is divided in the same manner as Mercator's Meridian, and corresponds with the table of meridional parts; except that the numbers in the latter are minutes, while the divisions on the other are degrees. Directly beneath the line of meridional parts, is placed a line of equal parts. The divisions of the latter being considered as degrees of longitude, the divisions of the former will be degrees of latitude adapted to the same scale. The meridional difference of latitude is found, by extending the compasses from one latitude to the other.

A chart may be constructed from the scale, by using the line of equal parts for the degrees of longitude, and the line of meridional parts for the intervals between the parallels of latitude.

85. It is an important property of Mercator's chart, that all the rhumb-lines projected on it are straight lines. This renders it, in several respects, more useful to navigators, than even the artificial globe. By Mercator's sailing, theorem II. (Art. 72.)

Merid. Diff. Lat. Diff. Lon.:: Rad. : Tan. Course.

So that, while the course remains the same, the ratio of the meridional difference of latitude to the difference of longitude is constant. If A, C, C', and C" (Fig. 26.) be several points in a rhumb-line AB, AB', and AB", the corresponding meridional differences of latitude, and BC, B'C', B'C', the differences of longitude; then

AB BC: AB' ; B'C' :: AB" : B"C". Therefore ABC, AB'C', AB''C'', are similar triangles, and ACC'C" is a right line. (Euc. 32. 6.)


86. The application of oblique angled trigonometry to the solution of certain problems in navigation, is called oblique sailing. It is principally used in bays and harbors, to determine the bearings of objects on shore, with their distances from the ship and from each other. A few examples will be sufficient here, in addition to those already given under heights and distances.

One of the cases which most frequently occurs, is that in which the distance of a ship from land is to be determined, when leaving a harbor to proceed to sea. This is necessary, that her difference of latitude and departure may be reckoned from a fixed point, whose latitude and longitude are known.

The distance from land is found, by taking the bearing of an object from the ship, then running a certain distance, and taking the bearing again. The course being observed, there will then be given the angles and one side of a triangle, to find either of the remaining sides.

Example I.

The point of land C, (Fig. 27.) is observed to bear N. 67° 30′ W. from A. The ship then sails S. 67° 30′ W. 9 miles from A to B ; and the direction of the point from B is found to be N. 11° 15' E. At what distance from land was the ship at A.

Let NS and N'S' be meridians passing through A and B. Then subtracting CAN and BAS each 671° from 180° we have the angle CAB-45°. And subtracting CBN/ 1110 from BAS or its equal ABN', we have ABC-561°. The angle at C is therefore 78° 45'. And

Sin. C AB:: Sin. B: AC=7.63 miles.

Example II.

New-York light-house on Sandy Point is in Lat. 40° 28' N. Lon. 74° 8' W. A ship observes this to bear N. 76° 16′ W., and after sailing S. 35° 10′ W. 8 miles, finds the bearing to be N. 17° 13' W. Required the latitude and longitude of the ship, at the first observation.

40° 261

The latitude is
The longitude

73 581

In this example, as the difference of latitude is small, the difference of longitude is best calculated by middle latitude sailing. (Art. 74.)

Example III.

A merchant ship sails from a certain port S. 51° E. at the rate of 8 miles an hour. A privateer leaving another port 7 miles N. E. of the first, sails at the rate of 10 miles an hour. What must be the course of the privateer, to meet the ship, without a change of direction in either?

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Example IV.

Two light-houses are observed from a ship sailing S. 38° W. at the rate of 5 miles an hour. The first bears N. 21° W., the other N. 47° W. At the end of two hours, the first is found to bear N. 5° E., the other N. 13° W. What is the distance of the light-houses from each other?

Ans. 6 miles and 30 rods.


87. When the measure given by the log-line is taken as the rate of the ship's progress, the water is supposed to be at rest. But if there is a tide or current, the log being thrown upon the water, and left at liberty, will move with it, in the same direction, and with the same velocity. The rate of sailing as measured by the log, is the motion through the


If the ship is steered in the direction of the current, her whole motion is equal to the rate given by the log, added to the rate of the current. But if the ship is steered in opposition to the current, her absolute motion is equal to the difference between the current, and the rate given by the log. In all other cases, the current will not only affect the velocity of the ship, but will change its direction.

Suppose that a river runs directly south, and that a boat in crossing it is steered before the wind, from west to east. It will be carried down the stream as fast, as if it were merely floating on the water in a calm. And it will reach the opposite side as soon, as if the surface of the river were at rest. But it will arrive at a different point of the shore.

Let AB (Fig. 28.) be the direction in which the boat is steered, and AD the distance which the stream runs, while the boat is crossing. If DC be parallel to AB, and BC parallel to AD; then will C be the point at which the boat proceeding from A, will strike the opposite shore, and AC will be the distance. For it is driven across by the wind, to the side BC, in the same time that it is carried down by the current, to the line DC.

In the same manner, if Am be any part of AB, and mn, be the corresponding progress of the stream, the distance sailed will be An. And if the velocity of the ship and of the stream continue uniform, Am is to mn, as AB to BC, so that

AnC is a straight line. (Euc. 32. 6.) The lines AB, BC, and AC, form the three sides of a triangle. Hence,

88 If the direction and rate of a ship's motion through the water, be represented by the position and length of one side of a triangle, and the direction and rate of the current, by a second side; the absolute direction and distance will be shown by the third side.

Example I.

If the breadth of a river running south (Fig. 28.) be 300 yards, and a boat steers S. 75° E. at the rate of 10 yards in a minute, while the progress of the stream is 24 yards in a minute; what is the actual course, and what distance must the boat go in crossing?

Cos. BAP AP ::R : AB=310.6

And 10 24 AB: BC=745.44.
Then in the triangle ABC,

(BC+AB): (BC-AB):: Tan. (BAC+BCA): Tan. (BAC—BCA)=17° ‍33′ 50′′.

Sin. BAC

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BC: Sin. ABC: AC=879 the distance.
And DAC-BCA=19° 56′ 10′′ the course.

Example II.

A boat moving through the water at the rate of five miles an hour, is endeavoring to make a certain point lying S. 224° W. while the tide is running S. 783° E. three miles an hour. In what direction must the boat be steered, to reach the point by a single course? Ans. S. 58° 33′ W.

89. But the most simple method of making the calculation for the effect of a current, in common cases, especially in resolving a traverse, is to consider the direction and rate of the current as an additional separate course and dis-tance; and to find the corresponding departure and difference of latitude. A boat sailing from A (Fig. 28.) by the united action of the wind and current, will arrive at the same point, as if it were first carried by the wind alone from A to B, and then by the current alone from B to C.

Example I.

A ship sails S. 17° E. for 2 hours, at the rate of 8 miles

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