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 reflexion, 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.

I observed the altitude of the top of a tower above the level sand on the sea-shore to be 59°; then, measuring directly from it 98 yards, its elevation was found to be 44°: required the height of the tower.

A

*The log. AC, by the preceding operation was found to be 2.42000, differing but a little from the log. of 23

PROBLEM X.

By observation, I found the angle of elevation of a monument, at one station, to be 21°, and the horizontal angle, at this station, between the spire of the monument and the second station, was 79; the horizontal angle, at the second station, between the spire and the first station, was 69°; the distance between the two stations being 139 yards: required the height of the monument.

Sailing towards the land, I discovered a 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 coluinn. 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; this process being similar to that in Problem VIIL 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 5.92 miles.

Distance

200 feet

18.71

... 24.63 statute miles, of about 694 to a degree; the distance in nautical leagues, of 20 to a degree, being about 7.

PROBLEM XII.

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 18.71

Distance

40 feet

8.37

27.08 miles.

PROBLEM XIII.

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 at the distance of a number of miles, 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 6 miles, because 30 divided by 4.6 gives 63 nearly.

PROBLEM XIV.

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 be at rest, and the wind blow 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 the apparent directions of the wind; the apparent being more ahead than the true, and the faster the vessel goes, the more ahead 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.

B

It is owing to the difference between the true and apparent 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 this, 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 the 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 equal to 7 points; the difference between the true and apparent directions of the wind will be obtained by drawing the line BC equal to 7 miles, taken from any scale of equal parts, and making the angle BCA equal to 7 points; then, with an extent equal to 27 miles, taken from the scale, and with one foot in B, describe an arc to cut the line AC in A; join AB; then the angle BAC, being measured, will be the required difference between the true and apparent directions of the wind.

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 will appear to change its directions above 24 points.

PROBLEM XV.

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 thermome ters, 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 velocity of sound at 32° Fahrenheit is 1090 feet per second, and for each additional degree of heat ad 406 to this velocity.

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 barometers 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 be 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 32°, otherwise subtracted: the sum or difference will be the true height of the upper above the lower observer in English fathoms, which, being multiplied by 6, will be the height in feet.

EXAMPLE.

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

MENSURATION.

PROBLEM I.

To find the area of a parallelogram.

RULE. Multiply the base by the perpendicular height; the product will be the area. Note. If both dimensions are given in feet, inches, &c., the product will be the area, expressed in square feet, square inches, &c., respectively. If one of the dimensions be given in feet and the other in inches, the product, divided by 12, will be the answer in square feet. If both dimensions are given in inches, the product will be square inches, which, being divided by 144, will be the answer in square feet. The same is to be understood in finding the area of other surfaces.

EXAMPLE I. Suppose the base BC of the rectangular parallelogram ABCD is 7 feet, and the perpendicular AB 3 feet; required the area. The product of the base 7 feet by the perpendicular 3 feet gives the area 21 square feet.

B

EXAMPLE II. Suppose ABCD is a board whose length BC is 22 feet, and breadth AB is 14 inches; required the number of square feet.

The product of the base 22 feet by the breadth 14 inches is 308; this, divided by 12, gives 253 square feet, the sought area.

EXAMPLE III. If BC be 25 inches, and AB 20 inches, required the area in square feet.

The product of the base 25 inches by the perpendicular 20 inches gives 500, which, divided by 144, gives the area 3.47 or 347 square feet.

EXAMPLE IV. Given the base AD of the oblique angular parallelogram ABCD, equal to 30 feet, and the perpendicular height BE 15 feet; required the area of the parallelogram. Multiply the base 30 feet by the perpendicular 15 feet; the product 450 is the area in square feet.

PROBLEM II.

To find the area of a triangle.

B

A

E

RULE. Multiply the base by half the perpendicular height, and the product will be the area required.

EXAMPLE. Given the base AC 30 feet, and the perpendicular BD 20 feet; required the area of the triangle.

The base 30 multiplied by half the perpendicular 10 gives the area 300 square feet.

PROBLEM III.

To find the area of any regular right-lined figure.

B

A

D

RULE. Reduce the figure to triangles, by drawing diagonals therein; then find the area of each triangle, and the sum of them will be the area of the proposed figure. Or, instead of finding the area of each triangle separately, you may find, at one operation, the area of two triangles, having the same diagonal, by multiplying the diagonal by half the sum of the perpendiculars let fall thereon.