2. If there be 10 panes of glass, each 4 feet 8 inches 9 parts long, and 1 foot 4 inches and 3 parts broad; how many feet of glass are contained in the 10 panes? Ans. 64·0407.

3.

There are 20 panes of glass, each 3 feet 6 inches 9 parts long, and 1 foot 3 inches and 3 parts broad; how many feet of glass are in the 20 panes ? Ans. 90-9224 ft.

4. If a window be 7 feet 6 inches high, and 3 feet 4 inches broad; how many square feet of glass contained therein? Ans. 25. 5. How many feet in an elliptical fan-light of 14 feet 6 inches in length, and 4 feet 9 inches in breadth?

Ans. 68 feet 10 inches. 6. What will the glazing of a triangular sky-light come to at 20d.; the base being 12 feet 6 inches, and the perpendicular height 6 feet 9 inches? Ans. £3 10s. 3d.

PAVERS' WORK.

Pavers' work is computed by the square yard; and the content is found by multiplying the length by the breadth. 1. What will be paid for paving a foot-path, at 4s. the yard, the length being 40 feet 6 inches, and the breadth 7 feet 3 inches?

2. What will be the expense of paving a rectangular courtyard, whose length is 62 feet 7 inches, and breadth 44 feet 3 inches; and in which there is a foot-path, whose whole

length is 62 feet 7 inches, and breadth 5 feet 6 inches, this at 38. per yard, and the rest at 2s. 6d per yard?

Ans. £39 118. 31d.

3. What is the expense of paving a court, at 3s. 2d. per yard; the length being 27 feet 10 inches, and the breadth 14 feet 9 inches? Ans. £7 4s. 54d.

4. What will the paving of a walk round a circular bowling-green come to, at 2s. 4d. per yard, the diameter of the bowling green being 40 feet, and the breadth of the walk 5 feet? Ans. £9 3s. 3d.

5. How many yards of paving in an elliptical walk 4 feet broad, the longer diameter being 60 feet, and shorter 50? Ans. 82-3797 yards.

VAULTED AND ARCHED ROOFS.

Arched roofs are either domes, vaults, saloons, or groins. Domes are formed of arches springing from a circular, or polygonal base, and meeting in a point directly over the centre of that base.

Saloons are made by arches connecting the side walls of a building to a flat roof, or ceiling.

Groins are made by the intersection of vaulted roofs with each other.

Vaulted roofs are sometimes circular, sometimes elliptical, and sometimes Gothic.

Circular roofs are those of which the arch is a part of the circumference of a circle.

Elliptical roofs are those of which the arch is a part of the circumference of an ellipsis.

Gothic roofs are made by the meeting of two equal circular arches, exactly above the span of the arch.

Groins are generally measured like a parallelopipedon, and the content is found by multiplying the length and breadth of the base by the height.

Sometimes one-tenth is deducted from the solidity thus found, and the remainder is reckoned as the solidity of the vacuity.

PROBLEM I.

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

RULE. Find the area of one end, by one of the foregoing rules, and multiply the area of the end by the length of the roof, or vault, and the product will be the content.

Note. When the arch is a segment of a circle, the area is found by Prob. XXVIII. Sec. II. When the arch is a segment of an ellipsis, multiply the span by the height, and that product by 7854 for the area of the end. When it is a Gothic arch, find the area of an isosceles triangle, whose base is equal to the span of the arch, and its sides equal to the two chords of the circular segment of the arch; then add the areas of the two segments to the area of the triangle, and the sum will give the area of the end.

1. What is the content of a concavity of a semi-circular vaulted roof, the span being 30 feet, and the length of the vault 150 feet?

30 x 30 900; then 900 x 7854

706.86, hence

2. What is the solid content of the vacuity AO EB of a Gothic vault, whose span A B is 60 feet, the chord B O, or A O, of each arch 60 feet; the distance of each arch from the middle of the chords as DE = 12 feet, and the length of the vault 40 feet?

DE3

123

DE) + = (60 × 12) + 60 x 2 = 4943 %

2 BO

In this example, the triangle A B O is equi-lateral, and its area is A B 3900√3 = 1557. Again, 3 (BOX =4943 = area of segment O E B, and 4943 × 2 = 988 the areas of the two segments OEB and OHA; then (1557 + 9883) × 40 101832 the solidity required.

Let MN KL represent a perpendicular section of a vaulted roof (Gothic). The span A B is 60 feet, the thickness of the wall M A, or B L, at the spring of the arch= 4 feet, the thickness OP at the crown of the arch=3, and the length of the roof 40 feet, the chord A O or O B = 60 feet, and the versed sine DE 12 feet; required the solidity of the materials of the arch.

✓ 602

=

First, (A O2 — A C2) = 302) 51.96 = SO the height of the vacuity of the arch, and SO+OP= 51.96354·96 = SP; again, A B+MA+BL= 60 + 4 + 4 = 68 = M L, and ML x S P = the area of the rectangle M NKL; hence, M L X SP × 40 101832, (the solidity of the vacuity A O B by the last Problem), gives the solidity of the materials; that is,

68 × 54.96 × 40 101832 = 47659-2 feet, the solidity required.

Note. When the arch A O B is an elliptical segment, its area multiplied by the length of the roof gives the solidity of the vacuity, and M L multiplied by S P, and the product by the length of the arch, gives the solidity of the cubic figure whose end is M N K L; and the difference of the two solidities is the solidity of the mixed solid whose section is AMNKL BEOHA. The materials of a bridge may be calculated after the same manner, by adding the solidities of T,T, and of the battlements, to the solidity as found in this Problem.

3. Required the capacity of the vacuity of an elliptical vault, whose span is 30 feet, and height 15 feet, the length of the vault being 90 feet. Ans. 31808-7 feet.

PROBLEM II.

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.

Note. To find the length of the arch, make a line ply close to it, quite across from side to side.

1. What is the surface of a vaulted roof, the length of the arch being 45 feet, and the length of the vault 140 feet?

140 x 45 6300 square feet.

2. Required the surface of a vaulted roof, the length of the arch being 40 feet 6 inches, and the length of the vault 100 feet? Ans. 4050 feet. 3. What is the surface of a vaulted roof, the length of the arch being 40.5 feet, and the length of the vault 60 feet? Ans. 2430 feet.

PROBLEM III.

To find the solidity of a dome, having the height and the dimensions of its base given.

RULE. Multiply the area of the base by the height, and of the product will give the solid content.*

This rule is correct only in one case, namely, when the dome is half a sphere, and in this case the height is equal to the radius of the circular base. It is a well-known property that the solidity of a sphere is 3 of that of a cylinder having the same base and height. But the solidity of a cylinder is found by multiplying the area of its base by the height. Hence the reason of the rule, when applied to this particular case. No general rule can be given to answer every case, as some domes are circular, some elliptical, some polygonal, &c.; they are of various heights, and their sides of different curvature. When the height of the dome is equal to the radius of its base, (the curved sides being circular, or elliptical quadrants,) or to half the mean proportional between the two axes of its elliptical base, the above rule will answer pretty well; but with any other dimensions it ought not to be used.