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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 wina, that it appears to shift its direction by tacking ship; and if the difference of the direc tions 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.

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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, tiken 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 velocity of sound at 32° Fahrenheit is 1090 feet per second, and for each additional degree of heat add 48 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 froin 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 subtractir, if the thermometer be highest at the lowest station, otherwise adding the sun or difference will be the approximate height in English fathoms. Multiply this by he decimal 0.00244, and by the difference between the mean of the two altitudes the detached thermometers and 32°; the product will be a correction, to be added to he 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.

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To find the area of a parallelogram.

RULE. Multiply the base by the perpendicular height; the product will be the &es. 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.

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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.

A

B

PROBLEM III.

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

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. O 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.

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EXAMPLE. Required the area of the figure ABCDE, in which CE = 33 feet BE=22 feet, and the perpendicular AF = 13 feet, BG = 14 feet, and DH = 12 feet.

The diagonal BE, 22 feet, multiplied by half the perpendicular AF, 6.5 feet, gives the area of the triangle ABE, 143 square feet; and the diagonal CE, 33 feet, multiplied by half the sum of the perpendiculars BG, DH, 13 feet, gives the area of the figure BCDE, 429 feet; this, added to the triangle ABE, 143 feet, gives the whole area 572 square feet.

PROBLEM IV.

To find the area of a circle.

B

RULE. Multiply the square of the diameter of the circle by the quantity 0.7854, and you will have the sought area.

Note. Instead of multiplying by 0.7854, you may multiply by 11 and divide by 14; the quotient will be the area nearly. This quantity, 0.7854, represents the area of a circle whose diameter is 1; the circumference of the same circle being 3.1416 nearly. The proportion of the diameter to the circumference is expressed in whole numbers by the ratio of 7 to 22 nearly, or more exactly by 113 to 355.*

EXAMPLE. Required the area of a circle ABCD, whose diameter BD is 10.6 feet.

The diameter 10.6 multiplied by itself and by 0.7854 gives the sought area, 88.247544 square feet.

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PROBLEM V.

To find the area of an ellipsis.

RULE. Multiply the longest diameter by the least, and the product by 0.7854; thi last product will be the area required.

EXAMPLE. Required the area of an ellipsis ABCD, whose longest diameter AC is 12 feet, and the shortest diameter BD 10 feet.

The product of the two diameters is 12 × 10–120; this, multiplied by 0.7854, gives the sought area, 94.2480 square feet.

D

B

The area of a sector of a circle may be found by means of the whole area of the circle obtained in Problem IV., by saying, As 360 degrees is to the angle contained between the two legs of the sector, so is the whole area of the circle to the area of the

sector.

There are various regular solids. The most noted are the following:-(1.) A Cube, which is a figure bounded by six equal squares. (2.) A Parallelopiped, which is a solid terminated by six quadrilateral figures, of which the opposite ones are equal and parallel. (3.) A Cylinder, which is a figure formed by the revolution of a rectangular parallelogram about one of its sides. (4.) A Pyramid, which is a solid decreasing gradually from the base till it comes to a point. There are various kinds of pyramids, according to the figure of their bases. Thus, if the base be a triangle, the solid is called a triangular pyramid; if a parallelogram, a parallelogramic pyramid; and if a circle, a circular pyramid, or simply a cone. The point in which the pyramid ends is called the vertex, and a line drawn from the vertex perpendicular to the base is calle the height of the pyramid.

This ratio may be easily remembered by observing that, if the first three odd numbers, 1, 3, 5, are repeated twice, they will produce the quantity 113355; the three first figures of which make the first

PROBLEM VI.

To find the solidity of a cube.

RULE. Multiplying the length of a side of the cube by itself, and the product by the sune length, gives the solidity required; which will be expressed in cubic feet if the dimensions be given in feet, but in cubic inches if the dimensions be given in inches, &c.

EXAMPLE. If the side AB of the cube be 6.3 feet, it is required to determine the solidity.

The product of 6.3 by 6.3 is 39.69; this, multiplied again by 6.3, gives the solidity 250.047 cubic feet.

D

PROBLEM VII.

A

To find the solidity of a rectangular parallelopiped.

RULE. Multiply the length, breadth, and depth, into each other; the product will be the solidity required.

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RULE. Multiply the square of the diameter of the base by the length, and this product by the constant quantity 0.7854; the last product will be the solidity required.

EXAMPLE. Required the solidity of a cylinder ADHF,

whose length DH. is 13 feet, and diameter of the base AD 11 feet.

The diameter 11, multiplied by itself and by the length 13, gives 1573, which, being multiplied by 0.7854, gives the H solidity in cubic feet 1235.4342.

PROBLEM IX.

To find the solidity of a grindstone.

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Grindstones, in the form of cylinders, are sold by the stone of 24 inches diameter and 4 inches thick. The number of stones that any one contains, may be obtained by the following rule.

RULE. Multiply the square of the diameter in inches by the thickness in inches, and divide the product by 2304, and you will have the number of stones required.

EXAMPLE. Required the number of stones in a grindstone whose diameter is 36 inches, and thickness 8 inches.

The square of the diameter 36 is 1296, which, being multiplied by the thickness 8 gives 10368. This, divided by 2304, gives 4.5, or 4 stones, the solidity required.

This problem may be solved by means of the line of numbers on Gunter's Scale, in a very expeditious manner, by the following rule.

RULE. Extend from 48 to the diameter; that extent, turned over twice the same way, from the thickness, will reach to the number of stones required.

Thus, in the preceding example, the extent from 48 to the diameter 36, turned over twice, from the thickness 8, will reach to 4.5, or 44, which is the number of stones sought.

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