PROBLEM IV.

Two ships sail from the same port, 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 pass NESW, and let its centre A repres the port sailed from; draw the N. W. ne AB=30 miles, and the N. E. by N. line AC= 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.

B

W

BY LOGARITHMS (by Cases IV. and V. Ob. Trig.)

Between the N. W. line AB and the N. E. by N. line AC, there are 7 points-angle BAC, half the supplement of which to 180° is 50° 87}'=half sum of the angles C and B.

To the angle C=40° 45', add the course from C to A=33° 45', the sum is 740 30', which is the bearing of B from C, viz. 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?

As the angle CBD is 78° 44′ or 7 points nearly, and the course from B to D is E. by N. the course from B to С must be north. The course from D to B being W. by S. or W. 11° 15′ S. and the angle BDC=40° 45′ the bearing of C from D must be W. 29° 30' N. because 40° 45'-11° 15′=29° 30′.

PROBLEM VI.

Coasting along shore, I saw two headlands, the first bore from me N. E. the second E. N. E. after sailing E. by S. 10 miles, the first bore N. by E. and the second N. E. by N. Required the bearing of the two headlands from each other, and their distance?

Draw the compass NESW, and let its centre A represent the place of the ship at the first station; draw the E. by S. line AB=10 miles, and B will be. the place of the ship at the second station; draw the N. E. line AC, and the E. N. E. line AD; through the point B draw the lines BC, BD parallel to the N. by E. and N. E. by N. lines, and the points C and D where they inter- W sect the lines drawn from A to the same headlands will be the points representing them respectively; join the points C and D ;-then will CD be the

38

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NIE

distance of the two headlands, and a line drawn through A parallel to CD will represent the bearing of those places from each other on the compass.

In the triangle ABD, we have all the angles and the side AB to find BD. For the bearings of B and A from D, are S. W. by S. and W. S. W. the difference being 3 points BDA; and the bearings of B and D from A, are E. by S. ard E. N. E. * the difference being also 3 points, equal to the angle BAD; therefore the a le BAD BDA, and (by art. 39 Geom.) BD-AB-10 miles. If these angles had not been equal, we might have calculated the side BD in the same manner as BC.

Now in the triangle CBD we have BD=10, BC14.97, and the angle CBD=22° 30', for the bearings of C and D from B are N. by E. and N. E. by N. differing 2 points or 220 30'; hence we may find the other angles and side CD as in case IV. Obl. Trig.

Being 96 fathoms from the bottom of a tower, I found its altitude above the horizontal line drawn from my eye was 150 10'. Required the elevation above that line?

When an object, whose elevation above the horizon is to be determined, is at a very great distance, it will be necessary to notice the correction arising from the curvature of the earth and the refraction, and apply that correction to the height estimated by the above method. Thus if the angular elevation of a mountain whose base was more distant than the limit of the visible horizon, was observed by an instrument of reflection; the approximate height must first be obtained, as in the preceding example, and then the correction of that approximate height for the curvature of the earth, refraction, and dip, must be calculated by the following rule, and added to that height, the sum will be the true height above the level of the sea.

RULE. Find in Table X. the number of miles corresponding to the height of the observer above the level of the sea, and take the difference between that number and the distance of the mountain from the observer in statute miles; with that difference enter the same table and find the height in feet corresponding, which will be the correction to be added to the approximate height to obtain the true height of the mountain above the level of the sea., EXAMPLE. Suppose the distance was 32 statute miles (or 168960 feet) and the observed altitude 1° 2′, the observer being 18 feet above the level of the sea. Required the height of the mountain above the same level?

PROBLEM VIII.

Sailing towards Cape-Cod, I discovered the light-house just appearing in the horizon, my eye being elevated 20 feet above the sea; it is required to find the distance of the light-house, supposing it to be elevated 200 feet above the surface of the sea?

The solution of this problem depends on the uniform curvature of the sea, by means of which all terrestrial objects disappear at certain distances from the observer. These distances may be computed by means of Table X. in which the elevation in feet is given in one column, and the distance at which it is visible, is expressed in statute miles in the other column. If the place from which you view the object be elevated above the horizon, you must add together the distances corresponding to the height of the observer and the height of the object, the sum will be the greatest distance at which that object is visible from the observer.

In the present example the height of the observer was 20 feet, and the height of the object 200 feet.

In Table X. opposite 20 feet is

200 feet

Distance

5,92 miles.

18,71

24,63 statute miles of about 694 to a de

gree, the distance in nautical leagues, of 20 to a degree, being about 7.

PROBLEM IX.

A man being on the main-top-gallant-mast of a man of war, 200 feet above the water, sees a 100 gun ship she had engaged the day before, hull to; how far were those ships distant from each other?

A ship of 100 guns or a first rate man of war, is about 60 feet from the keel to the rails, from which deduct about 20, leaves 40 for the height of her quarter-deck above water. Now a ship is seen hull to when her upper works just appear.

In Table X. opposite 200 feet stand
40 feet

Distance
PROBLEM X.

18.71

8.37

27.08 miles.

Upon seeing the flash of a gun, I counted 30 seconds by a watch before I heard the report: How far was that gun from me, supposing that sound moves at the rate of 1142 feet per second?

The velocity of light is so great, that the seeing of any act done even a number of miles distance, is instantaneous; but by observation it is found that sound moves at the rate of 1142 feet per second, or about one statute mile in 4.6 seconds; consequently the number of seconds elapsed between seeing the flash and hearing the report, being divided by 4.6 will give the distance in statute miles. In the present example the distance was about 64 miles, because 30 divided by 4,6 quotes 6 nearly.

PROBLEM XI.

A

To find the difference between the true and apparent directions of the wind. Suppose that a ship moves in the direction CB from C to B, while the wind moves in its true direction from A to B, the effect on the ship will be the same as if she was at rest and the wind blew in the direction AC with a velocity represented by AC, the velocity of the ship being represented by BC. In this case the angle BAC will represent the difference between the true and apparent directions of the wind; the apparent being more a-head than the true, and the faster the vessel goes the more a-head the wind will appear to be. We must, however, except the case where the wind is directly aft, in which case the direction is not altered.

It is owing to the difference between the true and appa

B

rent directions of the wind, that it appears to shift its direction by tacking ship; and if the difference of the directions be observed when on different boards (the wind on both tacks being supposed to remain constant, and the vessel to have the same velocity and to sail at the same distance from the wind) the half difference will be equal to the angle BAC; by knowing which, together with the velocity of the ship BC, and the angle BCA, we may obtain the true velocity of the wind; or, by knowing the velocity of the wind and of the ship, and the apparent direction of the wind, we may calculate the difference between the true and apparent directions of the wind.

Thus if the velocity of a ship represented by BC be 7 miles per hour, that of the wind represented by AB 27 miles per hour, and the angle of the vessel's course with the apparent direction of the wind BCA=7 points; the difference between the true and apparent directions of the wind would be obtained by drawing the line BC=7 miles taken from any scale of equal parts and making the angle BCA=74 points, then with an extent equal to 27 miles, taken from the scale, and with one foot in B describe an arch to cut the line AC in A, join AB ; then the angle BAC being measured, will be the sought difference between the true and apparent directions of the wind. BY LOGARITHMS.

So that in this case the difference between the true and apparent directions of the wind is about 14 points, and by tacking ship and sailing on the other board as above mentioned, the wind would appear to change its directions above 24 points.

PROBLEM XII.

To measure the height of a mountain by means of the heights of two barometers taken at the top and bottom of the mountain.

Procure two barometers with a thermometer attached to each of them, in order to ascertain the temperature of the mercury in the barometers, and two other thermometers of the same kind to ascertain the temperature of the air. Then one observer at the top of the mountain, and another at the bottom, must observe at the same time the heights of the barometers and the thermometers attached thereto, and the heights of the detached thermometers, placed in the open air, but sheltered from the sun. Having taken these observations, the height of the upper observer above the lower may be determined by the following rule, which is adapted to a scale of English inches and to Fahrenheit's thermometer.

RULE. Take the difference of the logarithms of the observed heights of the barometer at the two stations, considering the first four figures, exclusive of the index, as whole numbers, the remainder as decimals; to this difference must be applied the product of the decimal 0,454, by the difference of the altitudes of the two attached thermometers, by subtracting if the thermometer was highest at the lowest station, otherwise adding: the sum or difference will be the approximate height in English fathoms. Multiply this by the decimal 0,00244, and by the difference between the mean of the two altitudes of the detached thermometers and 32°, the product will be a correction to be added to the approximate height when the mean altitude of the two detached thermometers exceeds 320, otherwise subtracted; the sum or difference will be the true height of the upper above the lower observer in English fathoms, which multiplied by 6 will be the height in feet.

EXAMPLE.

Suppose the following observations were taken at the top and bottom of a mountain. Required its height in fathoms ?