Note. Balbec mentioned in the 6th question, is a town of Syria, situated about 37 miles north of Damascus. Its remains of antiquity display, according to the best judges, the boldest plan that ever was attempted.

The inhabitants of Asia consider Solomon as its founder; and the nobleness of the architecture, the beauty of the ornaments, and the stupendous execution of the whole, seem to fix its foundation to a period before the Christian æra; but, in all probability, the Jews knew little of the Grecian style of building and ornamenting in the time of Solomon.

PROBLEM IV.

To find the solidity of a cylinder.

RULE.

Multiply the area of the base by the perpendicular altitude of the cylinder, and the product will be the solidity.

Note 1. If the circumference of a cylinder be multiplied by its height, the product will be the convex surface; to which add twice the area of the base, and the sum will be the whole surface of the cylinder.

2. The altitude of a cylinder may be found by dividing the solidity by the area of the base.

3. The area of the base may be found by Problem 15, Part II.

EXAMPLES.

1. Required the solidity of the cylinder ABCD, the diameter of the base AB of which is 2.76 feet, and the altitude BC 4.58 feet.

Here 2.762 x .7854 2.76 x 2.76 x .7854 = 7.6176 x .7854 = 5.98286304, the area of the base; and 5.98286304 x 4.58 27.4015127232 feet, the solidity required.

2. The altitude of a cylindrical stone column is 8 feet 3 inches, and its circumference 5 feet 6 inches; what is its solidity? Ans. 19.86018 feet.

3. The circumference of a cylindrical piece of timber is 6 feet 8 inches, and its length 24 feet; what is its solid content? Ans. 84.88533 feet. 4. The diameter of a rolling-stone is 18.7 inches, and its length 4 feet 9 inches; required its solidity. Ans. 9.05952 feet. 5. The diameter of a well is 3 feet 9 inches, and its depth 45 feet; what did it cost sinking at 7s. 3d. per cubic yard? Ans. £.6 13s. 51d.

6. The greater diameter of a hollow iron roller is 1 foot 9 inches, the thickness of the metal 13 inch, and the length of the roller 5 feet; now supposing a cubic foot of cast iron to weigh 464 pounds avoirdupois; what did the roller cost at 19s. 9d. per hundred weight, and how many times will it turn round in rolling 5 acres of land?

Ans. The roller cost 13l. 1s. 0ğd. and it will turn round 7923.16926 times in rolling 5 acres of land.

PROBLEM V.

To find the solidity of a pyramid.
RULE.

Multiply the area of the base by of the perpendicular height, and the product will be the solidity.

Note. The surface of a pyramid may be found thus: Multiply the perimeter of the base by the length of the side, or slant height, and half the product will be the surface of the sides; to which add the area of the base, and the sum will be the whole surface.

2. The perpendicular height of a pyramid or cone is a line drawn from the vertex to the middle or centre of the base; and the slant height is the distance between the middle of one side of the base and the vertex.

3. In order to find the slant height of a pyramid from the perpendicular height, or the perpendicular height from the slant height, it will be necessary to find the distance between the centre of the base and the middle of one of its sides, which may be done by the following Rule: Multiply the number answering to the polygonal base, in the third column of the Table of polygons, Prob. 10, Part II., by the side of the base; and the product will be the distance required.

4. The area of the base of a pyramid may be found by Problems 11 and 12, Part 11.

EXAMPLES.

1. Required the solidity of the hexagonal pyramid AB CDE, each side of its base being 3 feet 9 inches, and its perpendicular altitude EF 15 feet 9 inches.

By Prob. 12, Part II., we have 2.5980762 × 3.752= 2.5980762 × 14.0625 = 36.5354465625, the area of the base; and 36.5354465625 × = 36.5354465625 ×

5.25

15.75
3

191.811094453125 feet, the solidity required. 2. The three sides of a triangular pyramid are 6, 7, and 8 feet, and its perpendicular altitude 18 feet; what is its solidity? Ans. 121.998975 feet. 3. The altitude of a square pyramid is 15 feet 9 inches, and the side of its base 2 feet 6 inches: required its solidity. Ans. 32 ft. 9 in. 9 pa.

4. The spire of a church is a regular hexagonal pyramid, whose perpendicular height measures 48 feet, the side of its base 10 feet, the perpendicular height of the cavity or hollow part 42 feet, and its side at the base 8 feet; how many cubic feet of stone are contained in the spire? Ans. 1829.0456448 feet.

5. The slant height of a regular, pentagonal, stone pyramid, measures 18 feet 6 inches, and the side of its base 3 feet; what was its whole expense; 6s. 9d. being paid for each cubic foot, and 64d. per square foot, for polishing the sides? Ans. £35 15s. 7 d.

PROBLEM VI.

To find the solidity of the frustum of a pyramid. GENERAL RULE.

To the areas of the two ends of the frustum add the square root of their product; and this sum being mul

tiplied by of the perpendicular height, will give the solidity.

Note. This rule will also serve for the frustum of a cone.

RULE II.

If the ends be regular polygons.

Add together the square of a side of each end, and the product of those sides; multiply the sum by of the height, and this product by the tabular area belonging to the polygon, (Prob. 12, Part II.) and the last product will be the solidity.

Note. The surface of the frustum of a pyramid may be found thus: Multiply half the sum of the perimeters of the two ends, by the slant height, and the product will be the surface of the sides; to which add the areas of the ends, and the sum will be the whole surface of the frustum.

EXAMPLES.

1. What is the solidity of the frustum ABCDEF, of a square pyramid; the side AB of the greater end being 5 feet, the side DE of the less end 2 feet, and the perpendicular height GH 12 feet?

C

Here 5 x525, the area of the greater end; and 2 × 24, the area of the less; also 25 x 4 =

100

10, the square root of their product; then 25+4+10 × 4 = 39 X 4 = 156 feet, the solidity required.

By Rule II.

Here 52 +22 + 5 x 2 = 25+ 4+10= 39; then 39 x 4 x 1156 feet, the solidity as before.

2. Each side of the greater end of a piece of squared timber is 28 inches, each side of the less end 14 inches, and its length 18 feet 9 inches; how many solid feet does it contain? Ans. 59.548611 feet.

3. The length of a piece of timber is 17 feet 3 inches, and its ends are similar rectangles; the length of the greater being 36, and its breadth 20 inches; the length of the less 18, and its breadth 10 inches; how many solid feet are contained in the frustum ?

Ans. 50.3125 feet. 4. What is the solidity of an octagonal stone pillar; each side of the greater end being 12 inches, each side of the less end 8 inches, and the perpendicular altitude 9 feet? Ans. 30.58003 feet.

5. The roof of a portico is supported by four regular, hexagonal, marble pillars, whose perpendicular altitude is 15 feet, the side of the greater end 8 inches, and the side of the less end 5 inches; what did they cost at 41. 10s. per solid foot? Ans. £209 9s. 42d.

6. The perpendicular altitude of a square chimney, in Leeds, is 120 feet 9 inches, the side of its base 10 feet 9 inches, and the side of its top 5 feet 9 inches. The cavity or hollow part is a square prism, whose side is 2 feet 6 inches; how many solid feet are contained in the chimney, and what is the area of its four sides? Ans. The solidity of the chimney is 7715.421875 cubic feet, and the area of the four sides is 3985.603941 square feet.

PROBLEM VII.

To find the solidity of a cone.

RULE.

Multiply the area of the base by the perpendicular height, and of the product will be the solidity.

Note 1. The altitude of an oblique cone, or one whose axis does not make a right angle with the plane of a base, may be found by demitting a perpendicular from the vertex of the cone, upon that plane.

2. If the circumference of the base of a cone be multiplied by half its slant height, the product will be the convex surface; to which add the area of the base, and the sum will be the whole surface of the cone.