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is to each of the other products, so is the weight of the compound to the weights of each of the ingredients.'

Let W, w, w', denote the weights of the compound and of the ingredients; and S, s, s', their specific gravities respectively, s being that of the denser ingredient; then

(s — s′) S : (S— s′) s =W : w, (s — s′) S : (s — S) s'—W: w

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EXAMPLE.-A composition weighing 56 lbs., having a specific gravity of 8.784, consists of tin and copper of the specific gravities 7.32 and 9 respectively; what are the quantities of the ingredients?

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13.176 14-75712×56=50

and hence w' = W w=56-50 = 6.

Or there are 56 lbs. of copper and 6 of tin.

By 72, the weight of a quantity of water, whose volume is equal to that of the body whose weight is w, is

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= ; and the same applies to the bodies whose weights

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and from these two equations are easily found the two formulas given above, and hence the origin of the rule.

EXERCISES.

1. An alloy of the specific gravity 7-8 weighs 10 lbs., and is composed of copper and zinc of the specific gravities 9 and 7-2; what is the weight of the ingredients?

Ans. 3.85 lbs. of copper, and 6·15 lbs. of zinc. 2. An alloy of the specific gravity 7.7 consisting of copper and tin of the specific gravities 9 and 7.3, weighs 25 ounces; what is the weight of each of the ingredients?

Ans. 6-87 ounces of copper, and 18·12 of tin.

ARCHED ROOFS.

75. Arched roofs are either vaults, domes, saloons, or groins.

Vaulted roofs consist of two similar arches springing from two opposite walls, and meeting in a line at the top, or else forming a continuous arch.

Domes are formed by arches springing from a circular or polygonal base, and meeting in a point above.

Saloons are formed by arches connecting the side walls with a horizontal roof or ceiling.

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

76. Arched roofs are either circular, elliptical, or gothic. In the first kind, the arch is a portion of the circumference of a circle; in the second, it is a portion of the circumference of an ellipse; and in the third kind, there are two arches which are portions of circles having different centres, and which meet at an angle in a line directly over the middle of the breadth, or span, of the arch.

77. By the cubic content of arched roofs is to be understood the content of the vacant space contained by its arches, and a horizontal plane passing through the base of the roof.

VAULTS.

78. PROBLEM I.-To find the cubic content of a vaulted roof.

'Multiply the area of one end, or of a vertical section, by the length.

=

Let a the area of the end, the length, and v = the volume, then v = al.

The areas of the ends are to be found by means of the rules in the Mensuration of Surfaces.

EXAMPLE. Find the volume of a semicircular vault, the span of which is 20, and its length 60 feet.

and

157·08,

a='7854 × 202 ×
val 157·08 × 60 = 9424.8.

EXERCISES.

1. Find the cubic content of an elliptic vault, whose span is 30, height 12, and length 60 feet. Ans. 16964-64. 2. What is the cubic content of a gothic vault, its span being 24, the chord of each arch 24, and the distance of each arch from the middle of its chord 9, and the length of the vault 30 feet? Ans. 17028-5.

79. PROBLEM II.-To find the surface of a vaulted roof. 'Multiply the length of the arch by the length of the vault."

Let a the length of the arch, that of the vault, and s = the surface, then s = al.

the

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EXAMPLE. What is the surface of a semicircular vault, span of which is 20, and length 60?

and

a=πr=3·1416 × 10=31·416,

sal=31·416 × 60 — 1884-96.

EXERCISES.

1. What is the surface of a circular vaulted roof, the span of which is 60 feet and its length 120 feet? Ans. 11309.8.

2. Find the surface of a vaulted roof, its length and that of its arch being 106 and 42.4 feet. Ans. 499-38 yards.

DOMES.

80. A dome with a polygonal base and circular arches, whose radii are equal to the apothem of the base, are called polygonal spherical domes.

81. PROBLEM III.-To find the cubic contents of a dome. Multiply the area of the base by two-thirds of the height.'

Let b = the base, h = the height, then the volume v = bh.

EXAMPLE. What is the solidity of a hexagonal spherical dome, a side of its base being 20 feet?

b = 4 × 6sh = 3 × 20 × h=60 h (art. 270) and h2=82-18282, for (fig. in art. 268, Part I.) ACB is in this case an equilateral triangle, and AC=s, AF = 1 s, and CF = h, also CF2 = AC2. AF2 (art. 204, Part I.) Hence, v=bh= × 60h × h = 40 h2 = 40 × 2 s2 = 30 × 20230 × 400=12000.

=

EXERCISES.

1. Find the content of a spherical dome, whose circular base has a diameter of 30 feet.

Ans. 7068.6. 2. What is the content of an octagonal dome, each side of its base being 40 feet, and its height 42 feet?

Ans. 216313.53.

82. PROBLEM IV.-To find the surface of a dome.

"When the dome is hemispherical, its surface is twice the area of the base.

8 = 2 × 7854 d2.

When the dome is elliptical on a circular base, add the height to half the diameter of the base, and multiply the sum by 1.5708 for the surface nearly.

8=1·5708 (d+h.)

In other cases, multiply double the area of the base by the height of the dome, and divide the product by the radius of the base for an approximation to the surface."

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EXAMPLE. Find the surface of a hexagonal spherical dome, each side of its base being 30 feet.

Here hr, and

s=2b=2× 302 × 2·598=4676.5.

EXERCISES.

1. How many square yards of painting are contained in a hemispherical dome of 50 feet diameter?

Ans. 436.3.

2. Find the surface of a dome with a circular base of 100

feet circumference, its height being 20 feet.

Ans. 2000.

SALOONS.

83. The vacuity of a saloon is the space contained by a horizontal plane through the base of the arches, the flat ceiling, and the arches.

84. PROBLEM V.-To find the vacuity of a saloon.

"Find the continued product of the height of the arc, its breadth or horizontal projection, of the perimeter of the ceiling and 3-1416.

From a side of the room, or its diameter when circular, take a like side or diameter of the ceiling, multiply the square of the remainder by the corresponding tabular area for regular polygons, or by 1 when the room is rectangular, or by 7854 when circular, and multiply this product by of the height.

Multiply the area of the flat ceiling by the height of the arch, and the sum of this product, and the two preceding, will be the content."

Let h, b, and p, be the height and breadth of the arc and perimeter of ceiling; S, s, two corresponding sides of the room and ceiling; a, a', the areas of the ceiling and of corresponding tabular polygon (271, Part I.); and A, B, C, the three products; then

A=7854 bhp, B= (S-s) 'a'h
Cah, and v=A+B+C.

For a square or rectangular room take 1 for a', and for a circular room take ·7854.

EXAMPLE. Find the cubic content of a saloon formed by a circular quadrantal arc of 2 feet radius, connecting a ceiling with a rectangular room 20 feet long and 16 wide. A='7854 bhp =7854 × 2 × 2 × 56 = 175.93 B=(S-8)'a'h (2016)2 x 1 x 2 = 21:33 = 16 x 12 x 2

C=ah

= 384.

=581.26

Hence,

v=A+B+C

EXERCISE. A circular building 40 feet diameter, and 25 feet high to the ceiling, is covered with a saloon, the circular

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