MENSURATION OF SOLIDS.

By the Mensuration of Solids are determined the spaces included by contiguous surfaces, and the sum of the measures of these including surfaces, is the whole surface or superficies of the body.

The measure of a solid, is called its solidity, capacity, or content.

Solids are measured by cubes, whose sides are inches, or feet, or yards, &c. And hence the solidity of a body is said to be so many cubic inches, feet, yards, &c., as will fill its capacity or space, or another of equal magnitude.

The least solid measure is the cubic inch, other cubes being taken from it according to the proportion in the following table:

Multiply the perimeter of one end of the prism by the length or height of the solid, and the product will be the surface of all its sides. To which, add also the area of the two ends of the prism, when required.†

Or, compute the areas of all the sides and ends separately, and add them all together.

Ex. 1.—To find the surface of a cube, the length of each side being 20 feet. Ans. 2400 feet. Ex. 2.-To find the whole surface of a triangular prism, whose length is 20 feet, and each side of its end or base 18 inches.

Ex. 3.—To find the convex surface of a round prism, length is 20 feet, and diameter of its base is 2 feet.

Ans. 91.948 feet. or cylinder, whose Ans. 125.664.

Ex. 4. What must be paid for lining a rectangular cistern with lead, at 2d. a pound weight, the thickness of the lead being such as to weigh 7 lb. for each square foot of surface; the inside dimensions of the cistern being as follow, viz. the length 3 feet 2 inches, the breadth 2 feet 8 inches, and depth 2 feet 6 inches? Ans. £2. 3s. 10d.

* Before perusing this chapter the student must make himself master of the treatise on the "Geometry of Solids," which immediately follows the "Geometry of Planes." The principle upon which the rules are founded are explained in the Differential Calculus.

+ The truth of this will easily appear, by considering that the sides of any prism are parallelograms, whose common length is the same as the length of the solid, and their breadths taken all together make up the perimeter of the ends of the same.

And the rule is evidently the same for the surface of a cylinder.

PROBLEM IL

To find the surface of a pyramid or cone.

Multiply the perimeter of the base by the slant height, or length of the side, and half the product will evidently be the surface of the sides, or the sum of the areas of all the triangles which form it. To which, add the area of the end or base, if requisite.

Ex. 1.-What is the upright surface of a triangular pyramid, the slant height being 20 feet, and each side of the base 3 feet? Ans. 90 feet. Ex. 2.-Required the convex surface of a cone, or circular pyramid, the slant height being 50 feet, and the diameter of its base 8 feet. Ans. 667.59.

PROBLEM II.

To find the surface of the frustum of a pyramid or cone; being the lower part, when the top is cut off by a plane parallel to the base.

Add together the perimeters of the two ends, and multiply their sum by the slant height, taking half the product for the answer. As is evident, because the

sides of the solid are trapezoids, having the opposite sides parallel.

Ex. 1.—How many square feet are in the surface of the frustum of a square pyramid, whose slant height is 10 feet; also, each side of the base or greater end being 3 feet 4 inches, and each side of the less end 2 feet 2 inches?

Ans. 110 feet. Ex. 2. To find the convex surface of the frustum of a cone, the slant height of the frustum being 12 feet, and the circumferences of the two ends 6 and 8.4. Ans. 90 feet.

PROBLEM IV

To find the solid content of any prism or cylinder.

Find the area of the base, or end, whatever the figure of it may be; and multiply it by the length of the prism or cylinder, for the solid content. Ex. 1.-To find the solid content of a cube, whose side is 24 inches.

Ans. 13824. Ex. 2.-How many cubic feet are in a block of marble, its length being 3 feet 2 inches, breadth 2 feet 8 inches, and thickness 2 feet 6 inches?

Ans. 21 Ex. 3.-How many gallons of water will the cistern contain, whose dimensions are the same as in the last example, when 277-274 cubic inches are contained in one gallon? Ans. 131.566.

Ex. 4.-Required the solidity of a triangular prism, whose length is 10 feet, and the three sides of its triangular end or base, are 3, 4, 5 feet. Ans. 60. Ex. 5.—Required the content of a round pillar, or cylinder, whose length is 20 feet, and circumference 5 feet 6 inches. Ans. 48.1459.

PROBLEM V.

To find the content of any pyramid or cone.

Find the area of the base, and multiply that area by the perpendicular height; then take of the product for the content.

Ex. 1.-Required the solidity of the square pyramid, each side of its base being 30, and its perpendicular height 25.

Ans. 7500. Ex. 2. To find the content of a triangular pyramid, whose perpendicular height is 30, and each side of the base 3. Ans. 38.97117.

Ex. 3. To find the content of a triangular pyramid, its height being 14 feet 6 inches, and the three sides of its base 5, 6, 7. Ans. 71-0352. Ex. 4. What is the content of a pentagonal pyramid, its height being 12 feet, and each side of its base 2 feet? Ans. 27.5276. Ex. 5.-What is the content of the hexagonal pyramid, whose height is 6-4, and each side of its base 6 inches? Ans. 1.38564 feet. Ex. 6. Required the content of a cone, its height being 10 feet, and the circumference of its base 9 feet. Ans. 22.56093.

PROBLEM VI.

To find the solidity of the frustum of a cone or pyramid.

Add into one sum, the areas of the two ends, and the mean proportional between them, or the square root of their product; and of that sum will be a mean area; which, being multiplied by the perpendicular height or length of the frustum, will give its content.

Ex. 1.-To find the number of solid feet in a piece of timber, whose bases are squares, each side of the greater en being 15 inches, and each side of the less end 6 inches; also, the length or perpendicular altitude 24 feet?

Ans. 19.

Ex. 2.-Required the content of a pentagonal frustum, whose height is 5 feet, each side of the base 18 inches, and each side of the top or less end 6 inches. Ans. 9-31925 feet.

Ex. 3. To find the content of a conic frustum, the altitude being 18, the greatest diameter 8, and the least diameter 4. Ans. 527-7888. Ex. 4. What is the solidity of the frustum of a cone, the altitude being 25, also the circumference at the greater end being 20, and at the less end 10?

Ans. 464.216.

Ex. 5.—If a cask, which is two equal conic frustums joined together at the bases, have its bung diameter 28 inches, the head diameter 20 inches, and length 40 inches; how many gallons of wine will it hold? Ans. 79.0613.

PROBLEM VIL

To find the surface of a sphere, or any segment.

RULE 1.-Multiply the circumference of the sphere by its diameter, and the product will be the whole surface of it.

RULE II.-Multiply the square of the diameter by 31416, and the product will be the surface.

Note. For the surface of a segment or frustum, multiply the whole circumference by the height of the part required.

Ex. 1.-Required the convex superficies of a sphere, whose diameter is 7, and circumference 22.

Ans. 154.

Ex. 2.-Required the superficies of a glube, whose diameter is 24 inches. Ans. 1809.5616.

Ex. 3.-Required the area of the whole surface of the earth, its diameter being 7957 miles, and its circumference 25000 miles.

Ans. 198943750 sq. miles. Ex. 4.-The axis of a sphere being 42 inches, what is the convex superficies of the segment, whose height is 9 inches? Ans. 1187-5248 inches.

Ex. 5.-Required the convex surface of a spherical zone, whose breadth or height is 2 feet, and cut from a sphere of 12 feet diameter. Ans. 78.54 feet

PROBLEM VIIL

To find the solidity of a sphere or Globe.

RULE 1.—Multiply the surface by the diameter, and take of the product for the content.

RULE II. Multiply the cube of the diameter by the decimal 5236, for the

content.

Ex. 1.—To find the content of a sphere whose axis is 12. Ans. 904-7808. Ex. 2. To find the solid content of the globe of the earth, supposing its circumference to be 25000 miles. Ans. 263,857,437,760 miles.

PROBLEM IX.

To find the solid content of a spherical segment.

RULE 1. From three times the diameter of the sphere take double the height of the segment; then multiply the remainder by the square of the height and the product by the decimal 5236. for the content.

RULE II.-To three times the square of the radius of the segment's base, add the square of its height; then multiply the sum by the height, and the product by 5236, for the content.

Ex. 1.-To find the content of a spherical segment, of 2 feet in height, cut from a sphere of 8 feet in diameter. Ans. 41.888.

Ex. 2.—What is the solidity of the segment of a sphere, its height being 9, and the diameter of its base 20? Ans. 1795-4244.

Note. The general rules for measuring all sorts of figures having been now delivered, we may next proceed to apply them to the several practical uses in life, as follows.

LAND SURVEYING.

SECTION I.

DESCRIPTION AND USE OF THE INSTRUMENTS.

1. OF THE CHAIN.

6 6

I AND is measured with a chain, called Gunter's Chain, from its inventor, of 4 poles or 22 yards, or 66 feet in length. It consists of 100 equal links; and the length of each link is therefore of a yard, or of a foot, or 7·92 inches. Land is estimated in acres, roods, and perches. An acre is equal to 10 square chains, that is, 10 chains in length and one chain in breadth. Or it is 220 × 22 = 4840 square yards. Or it is 40 x 4 = 160 square poles. is 1000 × 100 = 1,000,000 square links. These being all the same quantity. Also, an acre is divided into four parts called roods, and a rood into 40 parts called perches, which are square poles, or the square of a pole of 5 yards long, or the square of 4 of a chain, or of 25 links, which is 625 square links. So that the divisions of land measure will be thus:

Or it

The length of lines, measured with a chain, are best set down in links as integers, every chain in length being 100 links; and not in chains and decimals. Therefore, after the content is found, it will be in square links; then cut off five of the figures on the right hand for decimals, and the rest will be These decimals are then multiplied by 4 for roods, and the decimals of these again by 40 for perches.

acres.

EXAM.-Suppose the length of a rectangular piece of ground be 792 links, and its breadth 385: to find the area in acres, roods, and perches.

This instrument consists of a plain rectangular board, of any convenient size the centre of which, when used, is fixed by means of screws to a three-legged stand, having a ball and socket, or other joint, at the top, by means of which, when the legs are fixed on the ground, the table is inclined in any direction.