Note 1.-For a cube, take the cube of its side; and for a parallelopiped, take the continued product of the length, breadth and thickness, for the content.

Note 2.-The solid content will be of the same name as the dimensions which are multiplied together for finding the same; that is, if the dimensions be inches, the solidity will be cubic inches; or, if the dimensions be feet, then the solidity will be cubic feet, &c. If you multiply the area of the base in square inches by the length of the prism in feet, and divide the product by 144, the quotient will be the solid content in cubic feet.

Ex. 1. What is the solid content of a cylinder, or round prism, whose diameter is 2 feet, and length 12 feet? 2x.7854-3.1416, area of the base.

Here

Then, 3.1416×12=37.6992 cub. feet, Ans. 2. If a cubical bin, whose length, breadth and depth, are each 4 feet 2 inches, be filled to the brim with corn, how many bushels will it contain?

4ft. 2in. 50 inches, and 50×50×50=125000 cubic inches, the capacity of the bin. Then, 125000÷2150.4=58.12+ bushels, Ans.

3. What is the solid content of a stick of hewn timber, which is 12 inches broad, 8 inches thick, and 30 feet long?

This is a parallelopiped; therefore, 12×8×30=2880, and 2880÷144-20 cub. ft., Ans.

4. If a piece of hewn timber be 22 inches square, and 36 feet long, how many cubic feet does it contain?

Ans. 121. 5. 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, and 5 feet. Ans. 60 cub. ft.

Note 3.-The superficial content of any prism may be found thus: Multiply the perimeter, or circumference of one end of the prism, by the length, and the product will be the area of all its sides; to which add also the areas of the two ends of the prism, if required. The area of the whole surface of any cube is equal to 6 times the square of the length of one of its sides. PROBLEM II. To find the solid

RULE. Find the area of the

content of any Pyramid. base, by the rule for the

figure, whether it be a triangle, a square, a polygon, or a circle; then multiply this area by the perpendicular height of the pyramid, and one-third of the product will be the solid content; or multiply one of those factors by onethird of the other, and the product will be the solidity required.

Ex. 1. If the perpendicular height of a triangular pyramid be 30 feet, and each side of its base 10 feet, what is its solid content?

Here, the base is a trigon, or an equilateral triangle; therefore, 10x10x.43301=43.301, the area of the base. Then, (43.301×30)÷3=433.01 cubic feet, the solidity, Ans. 2. The largest of the Egyptian pyramids is said to be 693 feet square at its base, and 499 feet high: Required its solid content. Ans. 79881417 cub. ft.

3. Required the solid content of a round stick of timber, which is 30 feet long, 18 inches in diameter at the greater end, and tapers to a point at the other.

Ans. 17.67+ cub. ft.

Note. The superficial content of any pyramid may be found thus: Multiply the perimeter of the base, or sum of all its sides, by the slant height of the pyramid, and half of the product will be the area of all the sides; to which add the area of the base, if required.

PROBLEM III.—To find the solid content of any Frustum of a Pyramid.

RULE I.—Add into one sum, the areas of the two ends and the square root of the product of those areas, and onethird of this sum will be the mean area of a section between the two bases; which multiply by the perpendicular height or length of the frustum, and the product will be the solid

content.

RULE II.-If it be the frustum of a cone; multiply the diameters of the two bases together; to the product add one-third of the square of the difference of the said diameters; then multiply the sum by .7854, and the product will be the mean area between the two bases; which being multiplied by the height or length of the frustum, will give the solidity.

Bb

When the ends or bases of the frustum are any regular right lined figures; then, multiply the side of the greater base by the side of the less; to the product add one-third of the square of the difference of those sides; multiply the sum by the proper multiplier for the figure, found in the table of multipliers for regular polygons, and the product will be the mean area between the bases; which being multiplied by the length of the frustum, will give the solidity.

Ex. 1. If a round log be 20 feet long, its diameter at the greater end 16 inches, and at the less end 10 inches; how many cubic feet does it contain?

This is the frustum of a cone, and its solidity is found by Rule I. as follows: The arca of the greater end is =16×16×.7854-201.0624 sq. inches; that of the less end is=10x10x.7854-78.54 sq. in.; and the square root of the product of these areas is v(201.0624×78.54)=125.664

Then, (201.0624+78.54+125.664)÷3-135.0888 sq. in., the mean area between the bases; and (135.0888×20)÷ 144-18.76+cub. feet, Ans.

The same answer may be found more easily by Rule II. 2. Required the solid content of a tapering stick of hewn timber, whose ends are squares; the greater end being 20 inches square, the less end 14 inches, and the length 30 feet.

By Rule II.-20-14-6, and (6×6)+3=12. Then, (20 ×14+12)x1=292 sq. inches, the mean area; and (292×30) +144-60 cubic feet, Ans.

Note. The superficial content of any frustum of a pyramid may be found thus: Add together the perimeters of the two ends; multiply the sum by the slant height, and half the product will be the area of the sides; to which add the areas of the two ends, if required.

PROBLEM IV. To find the area of the surface of a Sphere or Globe.

RULE I.-Multiply the circumference of the sphere by its diameter, and the product will be the area of its surface.

RULE II.-Multiply the square of the diameter of the sphere by 3.1416.

RULE III.-Multiply the square of the circumference by .31831

Ex. Supposing the earth to be a globe, whose diameter is 7930 miles, and circumference 24912 miles, what is the area of its surface?

The answer, found by Rule I. is 197552160 sq. miles. PRORLEM V. To find the solid content of a Sphere or Globe.

RULE I.-Multiply the area of the surface by the diameter of the sphere, and take one-sixth of the product, for the solidity: Or, which is the same thing, multiply the square of the diameter by the circumference, and take onesixth of the product.

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

RULE III.—Multiply the cube of the circumference by .016887

Ex. If the earth be a globe, whose diameter is 7930 miles, and circumference 24912 miles, how many cubie miles are contained in it?

The answer, found by Rule I., is 261,098,104,800 cubic miles.

PROBLEM VI. To find the solid content of any Segment of a Globe.

RULE. To three times the square of the semidiameter 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. A collier, wishing to burn a quantity of wood into charcoal, piled it up into a stack in the form of a segment of a globe, the height of which was 9 feet, and the diameter of the base 20 feet: how many cubic feet of wood did the pile contain? Ans. 1795.4244 cub. ft.* PROBLEM VII.-To find the solid content of a Spheroid.

* The solid content of any coal pit, of the usual form, may easily be found as follows: Measure with a rope or chain across the top of the stack of wood, from the bottom on one side to the bottom on the opposite side; then divide the cube of this measure, in feet, by 1895, and the quotient will be the required content, in cords.

RULE.-Multiply the square of the shortest diameter by the longest diameter; then multiply the product by .5236, and the last product will be the solidity.

Ex. If the length, or longest diameter of a spheroid, be 40 inches, and the shortest diameter 30 inches, what is its solidity? Ans. 18849.6 cub. inches. PROBLEM VIII.—To find the solid content of any irregular body.

RULE.-Put the body, or quantity which you would measure, into any vessel of regular shape, and then fill the vessel with water, sand, or any other convenient substance. Then take out the body, and measure the space left empty by its removal, according to the preceding rules, and the content of this space will be the solid content of the irregular body.

MENSURATION OF BOARDS AND TIMBER.

PROBLEM I.-To find the area or superficial content of any Board.

RULE.-Multiply the length of the board by the mean breadth, and the product will be the superficial content required.

Note 1.-When the board is broader at one end than at the other; if it tapers regularly, take the breadth in the middle; or else add the breadths of the two ends together, and take half the sum, for the mean breadth: But, if the board does not taper regularly, you may measure the breadth in several places at equal distances, and then divide the sum of these breadths by the number of them, and the quotient will be the mean breadth.

Note 2.-If you multiply the length in feet by the mean breadth in inches, and divide the product by 12, the quotient will be the superficial content in square feet.—In measuring boards, timber, &c. it will be convenient to make use of a measuring rule in which the feet are divided decimally, viz. each foot into 10 equal parts, and these again into 10 parts; for then the dimensions may be taken