OF

PAVIORS' WORK.

PAVIORS' work is done by the square yard, and the content is found by multiplying the length by the breadth.

Or if the dimensions be taken in feet, and the area be found in the same measure, the result being divided by 9, will give the number of square yards required.

EXAMPLES.

1. What will the paving of a rectangular-court-yard come to at 3s. 2d. per yard, supposing the length to be 27 feet 10 inches, and the breadth 14 feet 9 inches?

Plumbers' work is generally done by the pound, or hundred weight, and the price is regulated according to the value of the lead at the time the contract is made, or when the work is performed. Sheet lead, used in roofing, guttering, &c. is generally between 7 and 12lbs. weight to the square foot.

The following table will show the weight of a square foot to each of these thicknesses.

2. A rectangular court-yard is 42 feet 9 inches long, and 68 feet 6 inches in depth, and a foot-way goes quite through it, of 5 feet 6 inches in breadth; the foot-way is laid with stone at 3s. 6d. per yard, and the rest with pebbles at 3s. per yard: what will the whole come to? Ans. 491. 17s. 0‡ā.

R

OF

VAULTED AND ARCHED ROOFS.

ARCHED roofs are either vaults, domes, saloons, or groins. Vaulted roofs are formed by arches springing from the opposite walls, and meeting in a line at the top.

Domes are made by arches springing from a circular or polygonal base, and meeting in a point at the top.

Saloons are formed by arches connecting the side walls to a flat roof, or ceiling, in the middle.

Groins are formed by the intersection of vaults with each other.

Domes and saloons rarely occur in the practice of measuring, but vaults and groins over the cellars of most houses. Vaulted roofs are generally one of the three following

sorts:

1. Circular roofs, or those whose arch is some part of the circumference of a circle.

2. Elliptical roofs, or those whose arch is some part of the circumference of an ellipsis.

3. Gothic roofs, or those which are formed by two circular arcs that meet in a point directly over the middle of the breadth, or span of the arch.

PROBLEM I.

To find the solid content of a circular, elliptic, or gothic

vaulted roof.

RULE.*

Multiply the area of one end by the length of the roof, and the product will be the solidity required.

EXAMPLES.

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

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

Ans. 30159.36 feet.

3. What is the solid content of a gothic vault, whose span is 48, the chord of its arch 48, the distance of the arch from the middle of the chord 18, and the length of the vault 60? Ans. 136224.71712.†

PROBLEM II.

To find the concave, or convex surface, of circular, elliptic, or gothic vaulted roofs.

*To find the solidity of the materials in either of the arches. RULE. From the solid content of the whole arch take the solid content of the void space, and the remainder will be the solidity of the arch.

†The areas of these segments were calculated by rule 2, Prob. XIII. and the triangle by Prob. VIII.

RULE.*

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

EXAMPLE.

What is the concave surface of a semi-circular vault, whose span is 40 feet and its length 120?

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.

EXAMPLES.

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

* The convex surface of a vault may be found by stretching a string over it; but for the concave surface this method is not applicable, and therefore its length must be found from proper dimensions.

+ Domes and saloons are of various figures, but they are things that seldom occur in the practice of measuring.