2. The base of a triangle measures 76 feet 8 inches, and the perpendicular 42 feet 3 inches; what did it cost paving with Aberdeen granite, at 10s. 6d. per square yard? Ans. £94 9s. 6d.

3. If the parallel sides of a trapezoid be 68 feet 7 inches, and 45 feet 3 inches, and their perpendicular distance 98 feet 6 inches; what will it cost paving with Guernsey pebbles, at 6s. 6d. per square yard? Ans. £202 8s. 113d.

4. How many yards of paving does the trapezium contain, whose diagonal measures 136 feet 8 inches, and perpendicular 68 feet 2 inches, and 56 feet 4 inches? Ans. 945 yds. 21 ft 5. The length of a street is 538 feet 6 inches, and its breadth 65 feet 8 inches; what did it cost paving with Purbeck-stone, at 5s. 6d. per square yard?

Ans. £1080 9s. 94d.

6. A rectangular court-yard measures 96 feet 9 inches in length, and 74 feet 6 inches in breadth. Across the middle and round the extremities of the yard, is a footway, 5 feet 3 inches broad; and paved with Guernsey granite, at 9s. 6d. per square yard. The rest is paved with Jersey pebbles, at 5s. 9d. per square yard; required the expense of the whole. Ans. £272 8s. 32d.

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.

Vaulted roofs are commonly of the three following

sorts :

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

2. Elliptical or oval roofs, or those whose arch is an oval, or some part of the circumference of an ellipsis. 3. Gothic roofs, or those which are formed by two cir

cular arcs, struck from different centres, and meeting in a point over the middle of the breadth or span of the arch.

Note. Domes and saloons are of various figures; they, however, seldom occur in the practice of measuring; but most cellars are covered either with vaults or groins.

PROBLEM I.

To find the content of the vacuity of a circular, an elliptic, or a Gothic vaulted roof.

RULE.

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

Note 1. If the arch be the segment of a circle, the area of the end may be found by Problem 17, Part II.; if it be elliptical, multiply the span by the height, and the product by .7854, for the area of the end; but if it be a Gothic arch, the area of the end must be obtained by finding the areas of the two circular segments and the triangle of which the end is composed.

2. The upper sides of all arches, whether vaults or groins, are built up solid, above the haunches, to the same height as the crown of the arch.

3. The solidity of the materials in any arched roof, may be found thus: find the content of the whole, considered as solid, from the spring of the arch to the upper side of the crown; find also the contents of the vacuity; then the difference of these two contents will be the solidity required.

4. The whole arch, considered as a solid, will be a parallelopipedon, the content of which may be found by Problem 2, Part IV.

EXAMPLES.

1. Required the content of the vacuity of a semi-circular vault, the span or diameter of which is 20 feet, and its length 60 feet.

.7854

400 the square of 20.

2)314.1600

157.08 the area of the end.

60 the length.

9424.80 Ans.

2. The span of an elliptical vault is 30 feet, its height 10 feet, and its length 50 feet; what is the content of the vacuity? Ans. 11781 feet.

3. Required the content of the vacuity of a Gothic vault, whose span is 30 feet, the chord of each arch 32 feet, the versed sine, or distance of each arch from the middle of these chords, 8 feet, and the length of the vault 35 feet 6 inches. Ans. 27755.0502 feet.

4. Let ABCD denote the end or upright section of a semi-circular roof. The span EF is 18 feet, the thickness of the wall AE or FB, at the spring of the arch, 3 feet, the A thickness GH, at the crown of the arch, 2 feet, and the length of the

vault 56 feet 9 inches; how many solid feet are contained in the roof? Ans. 7761.425 feet.

PROBLEM II.

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

RULE.

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

Note. The convex length of an arch may be easily found by making a line ply close over it; but for the concave length, this method is not quite so applicable; for if care be not taken, the dimension will be made too short.

If the arch be the segment of a circle, its true length may be found by Problem 14, Part II.

EXAMPLES.

1. The span of a semi-circular vault is 30 feet, and its length 40 feet; what is its concave surface?

3.1416
30

2)94.2480

47.124 = length of the arch.

40

Ans. 1884.960 square feet.

2. The length of a vault is 62 feet 9 inches, and that of the arch 54 feet 6 inches; how many square yards are contained in the roof? Ans. 379.986 yards.

3. Required the concave surface of a bridge consisting of 5 circular arches; the span of each arch being 96 feet, the height, above the top of the piers, 36 feet, and the length 45 feet. Ans. 28955.232 feet.

Note. Those who desire to make themselves acquainted with the essential properties, dimensions, proportions, and other relations of the various parts of a bridge, are referred to Dr. Hutton's Principles of Bridges.

In this valuable little work, the learned Doctor proves, that the equilibrial arch, described in Problem V., is the most proper for a bridge of several arches. Next to it, the elliptical arch claims the preference; after it the cycloidal arch; and lastly the arch of a circle.

As for parabolic, hyperbolic, and catanarian arches, they ought never to be admitted into a bridge consisting of several arches; but may, in some cases, be used for a bridge of one arch, which is to rise an unusual height.

PROBLEM III.

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

RULE.

33

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

EXAMPLES.

1. What is the solid content of a hemispherical dome; the diameter of the base being 40 feet?

.7854

1600 = square of 40.

4712400

7854

1256.6400 area of the base.

20= height.

25132.8000

2

3)50265.6000

16755.2000 = solidity.

2. Required the solidity of an octagonal dome; each side of the base being 20 feet, and the height 21 feet.

PROBLEM IV.

Ans. 27039.19176 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 superficies required.

Note. If the dome be elliptical, the product of the two diameters multiplied by 1.5708, will give the superficial content, sufficiently near for practical purposes.

EXAMPLES.

1. Required the superficies of a hexagonal spherical dome; each side of the base being 20 feet.

2.5980762 = tabular area, Prob. 12, Part II. 400 square of 20.

1039.2304800 = area of the base.

2. What will an octagonal spherical dome cost painting, at 1s. 3d. per yard; each side of the base being 10 feet? Ans. £6 14s. 1d.

PROBLEM V.

To find the solid content or vacuity of a saloon.

RULE.

Multiply the area of a transverse section by the mean compass of the solid part of the saloon; subtract this product from the whole vacuity of the room, supposing the walls to go upright, all the height, to the flat ceiling, and the difference will be the answer

Note 1. If the base of the saloon be a rectangle, the vacuity of the room, or the whole upright space, will be a parallelopipedon, the content of which may be found by Problem 2: if the base be a regular polygon, the vacuity will be a prism, the content of which may be found by Problem 3; and if the base be a circle, the vacuity will be a cylinder, the content of which may be obtained by Problem 4, Part IV.

2. For practical purposes, the mean compass of the solid part may be found by adding the compass taken at the middle of the arch to the compass of the room, taken within the walls, and dividing the sum by 2; but in solving the second and third examples, the mean compass of the solid part was found with mathematical accuracy.

EXAMPLES.

1. If the height AB of a saloon be 3.2 feet, the chord ADC of its face 4.5 feet, and the distance DE of its middle part from the arch be 9 inches; required the solidity, supposing the mean compass of the saloon to be 50 feet.

By Rule 2, Problem 17, Part II., C we have (4.5x.75)x+.753+(4.5 x 2)=3.375 × ×.4218759=2.25 + .046875=2.296875, the area of the segment ADCEA.

Again, by Problem 6, Part II., we have AC-AB2=4.5° -3.2o=20.25 10.24 10.01; and 10.01 = 3.163858 BC; then BC× AB=

E

B

3.163858 × 1.6=5.0621728, the area of the triangle ABC;