that meet in a point directly over the middle of the breadth or span of the arch.

To find the solid content of a circular, elliptical, or Gothic vaulted roof.

RULE. Find the area of one end, and multiply it by the length of the roof; the product will be the solidity required.

1. Required the solidity of an elliptic vault whose span is 40 feet, height 12 feet, and length 80 feet.

Thus, 12 x 40 x 7854 376.992

area of one end. And 376-992 x 8030159-36 feet, solidity required.

2. What is the solid content of a semi-circular vault whose span is 40 feet, and its length 120 feet?

Thus, 40 × 40 = 1600 × ·7854 = 1256·64 ÷ 2

75398-40. Ans.

628.32.

3. What is the solidity of a semi-circular vault whose span is 40 feet, and length 124 feet?

Ans. 77911.68.

To find the concave or convex surface of a circular, elliptical, or Gothic vaulted roof.

RULE.-Multiply the length of the arch by the length of the vault, and the product will be the superficies required.

1. What is the surface of a vaulted roof, the length of the arch being 35 feet, and the length of the vault 140 feet?

=

Thus, 35 x 140 4900 square feet, the surface required. 2. Required the surface of a vaulted roof, the length of the arch 40 feet 6 inches, and the length of the vault 79 feet.

Ans. 3199.5 feet.

To find the solid content of a dome, its height and the dimensions of its base being known.

RULE.-Multiply the area of the base by two-thirds of the height, and the product will be the solidity.

1. What is the solid content of a spherical dome, the diameter of whose circular base is 60 feet, and height 30 feet?

Thus, 60 × 60

56548.8. Ans.

= 3600 × 7854 =2827·44 × 20 (3 of 30) =

=

2. What is the solid content of an octagonal dome, each side of its base being 20 feet, and the height 21 feet?

Ans. 27039.1912 feet.

To find the superficial content of a spherical dome.

RULE.-Multiply the area of the base by 2, and the product will be the superficial content required.

1. Required the superficial content of an hexagonal spherical dome, each side of the base being 20 feet.

=

Thus, 20 x 20 400 x 2.598076 = whose side is 1,) 1039-2304

Answer.

(area of a hexagon area of base × 2 2078-4608.

2. What will be the expense of painting an hexagonal spherical dome, each side of whose base is 20 feet, at 15 cents per square yard? Ans. D. 34.641.

To find the solid content of the vacuity formed by a groin or arch, either circular or elliptical.

1

RULE.-Multiply the area of the base by the height, and the product by 904, and it will give the solidity required.

1. What is the solid content of the vacuity formed by a circular groin, one side of its square base being 12 feet, and length 6 feet?

Thus, 122 x 6 x 904 781.056 solidity required.

=

2. What is the solid content of the vacuity formed by an elliptical groin, one side of its square base being 20 feet, and height 6 feet? Ans. 2169-6 feet.

To find the concave superficies of a circular groin. RULE.-Multiply the area of the base by 1.1416, and the product will be the superficies required.

1. What is the curve superficies of a circular groin arch, one side of its square being 12 feet?

=

Thus, 12 x 12 x 1·1416 164-3904, superficies required. 2. What is the concave superficies of a circular groin arch, one side of its square being 9 feet? Ans. 92-4696.

To find the superficies of a saloon.

RULE.-Find its breadth by applying a line close to it across the surface; find also its length by measuring along the middle of it, quite round the room; then multiply these two dimensions together for the superficial content.

1. The girt across the face of the saloon is 4 feet, and its mean compass 99 feet; ; what is the superficial content?

Thus, 99 × 4

396 feet, the superficial content. 2. The girt across the face of a saloon is 24 feet, and the mean compass 196 feet; required its surface.

3. The girt across the face of the saloon is 18 feet, its mean compass 184 feet; required its surface.

4. The girt across the face of the saloon is 28 feet, its mean compass 224 feet; required its surface.

SPECIFIC GRAVITY.

The specific gravity of bodies are their relative weights contained under the same given magnitude, as a cubic foot, a cubic inch, &c.; and are expressed by the numbers annexed to their names in the following table, as found by actual experiments, and are calculated to correspond with a cubic foot of each avoirdupois.

It has been ascertained that a cubic foot of rain water weighs 62 pounds, or 1000 ounces avoirdupois; and a cubic foot containing 1728 cubic inches, it follows that a cubic inch weighs 03616898148 of a pound; hence by multiplying the specific gravity of a body by the above number, the product will be the weight of a cubic inch of that body in pounds avoirdupois, which being multiplied by 175, and the product divided by 144, the quotient will be the weight of a cubic inch in pounds troy144 pounds avoirdupois being equal to 175 pounds troy.

NOTE. All bodies expand with heat and contract with cold, but some more and some less than others; consequently the specific gravity of bodies is not precisely the same in summer as in winter.

The specific gravity of a body, and its weight being given, to find its solidity.

RULE. As the tabular specific gravity of the body is to its weight in avoirdupois ounces, so is one cubic foot, or 1728 cubic inches, to its content in feet or inches respectively.