2

2x2 x.7854-4=3.1416=one fourth of product of the square of twice the height of the arc and .7854.

3.1416 × (20-2+2)×4=201.0624=product of preceding quotient and the length of the sides of the ceiling.

2+2-2016, and 16-20=4=diameter of ceiling subtracted from diameter of saloon.

42×1=16=product of square of side and tabular multiplier = =area of

figure.

16x of 2=21.333=product of area of base and of height of ceiling. 162×2=512 product of area of ceiling and its height.

512+21.333+201.0624=734.3954=sum of above products=volume

required.

Ex. 2. A circular room 40 feet diameter, and 25 feet high to the ceiling, is covered with a saloon having a circular arch of 5 feet radius; required the contents of the room in cubic feet. Ans. 30779.453 feet.

Operation.--Area of 40 × 25-5=1256.64×20=25132.8 = volume of body of room.

Area of flat portion of ceiling, 40-5+5=706.86×25 20=3534.3 =volume of body of roof.

78.54

Area of quadrant of a circle having a radius of 5= =3 =19.625. Volume of quadrantal ring of 5 feet height and base=area then of the circumference described by its centre of gravity. (See p. 209.) Hence (by rule, p. 84), the centre of gravity of the ring having the section of a sector of a circle of 5 feet radius is 3.001 feet from the angle of it. Then (by rule, p. 55), the hypothenuse of the right angle (3) being alone given, the length of the side (that is, the distance of the centre of gravity of the sector from the vertical side of it)=2.122. Therefore, 40−5+5 +2.122×2=34.244 diameter of circle described by centre of gravity of quadrantal ring.

Consequently, circumference of 34.244 × 19.635=2112.353=volume of quadrantal ring.

Volume of body of room,

25132.8

30779.453 cubic feet-volume required.

Ex. 3. A quadrangular building having sides of 40 and 30 feet is covered with a saloon 25 feet in height from the floor, having an arch of 5 feet radius; what is the volume of the saloon? Ans. 29296.83 cubic feet.

To ascertain the Surface of a Vault.

RULE.-Multiply the length of the arch by the length of the vault, and the product will give the surface required. Or, pxl=surface, p representing the perimeter of the arch.

EXAMPLE.-What is the concave or internal surface of a circular vault, the width of it being 40 feet and the length 80? 40×3.1416÷2×80=5026.56=product of length of arch and length of vault=result required.

Ex. 2. The width of an elliptic arched vault is 18 feet, its height 12, and its length 50; what is its internal surface? Ans. 1666.085 square feet.

To ascertain the Volume of a Vault.

RULE.-Multiply the area of a section of the vault by its length, and the product will give the volume required. Or, axl=volume.

EXAMPLE.-The width of a semi-circular arched vault is 10 feet, and its length 60; what is its volume?

102x.7854-2=39.27=area of semicircle of 10 feet span.
39.27×60=2356.2=product of area and length=volume required.

Ex. 2. The width of a semi-elliptic arched vault is 25 feet, its height 17.5, and its length 40 feet; what are its contents? Ans. 13744.5 cubic feet.

To ascertain the Internal Surface of a Circular Groin. RULE.-Multiply the area of the base by 1.1416, and the product will give the surface required.

Or, a× 1.1416=surface.

NOTE.-The exact surface or volume of a groin is obtained by subtracting from the sum of the surfaces or volumes of the two vaults composing the groin, the surface or volume of the quadrantal arch formed by them. Thus, in Example 1, the surface of a circular groin of 12 feet base is as follows:

=

Surface of vault, 12 × 3.1416÷2 18.8496, which x 12 and 2 452.3904 surface of the two vaults.

122x2=288=surface of the quadrantal arch. Hence, 452.3904-288-164.3904=surface.

==

EXAMPLE.—What is the surface of a circular groin, a side of its square being 12 feet.

122 × 1.1416=164.3904=product of area of base and 1.1416=surface required.

Ex. 2. What is the surface of a circular groin, a side of its square being 8 feet? Ans. 73.0624.

NOTE. This rule may be observed in elliptical groins, as the error or difference is too small to be regarded in ordinary practice.

To ascertain the Volume of a Circular or an Elliptical Groin. RULE.-Multiply the area of the base by the height, and the product again by .904, and it will give the volume required.

EXAMPLE.-What is the volume of the vacuity or space formed by a circular groin, one side of its square being 10 feet?

102×5X.904=452=product of area of base, the height and .904= volume required.

Ex. 2. What is the volume of the vacuity formed by an elliptic groin, one side of its square being 24 feet, and its height 9 feet? Ans. 4686.336 cubic feet.

To ascertain the Internal Surface of a Triangular Groin.

RULE.-Ascertain the length of a side of the arch, multiply it by twice the width of the vault, and the product will give the surface required.*

Or, lxbx2=surface.

EXAMPLE. The width of a triangular groin is 12 feet, and its height 12; what is its internal surface?

2

✓ (122+12÷÷2)=13.4164= length of one side of arch.

13.4164 × 12×2=321.9936=product of length of side and twice the base=result required.

*See note, page 273.

To ascertain the Volume of the Materials that form the Groin.

RULE.-Multiply the area of the base by the height, including the work to the top of the groin, and from this product subtract the volume of the vacuity; the result will give the volume required.

A General Rule for the Measurement of the Contents of Arches is thus:

From the volume of the whole, considered as a solid, from the springing of the arch to the outside of it, deduct the vacuity contained between the said springing and the under side of it, and the remainder will give the contents of the solid part.

In measuring works where there are many groins in a range, the cylindrical pieces between the groins, and on their sides, must be computed separately.

When the upper sides of orophoids, whether vaults or groins, are built up solid, above the haunches, to the height of the crown, it is evident that the product of the area of the base and the height will be the whole contents. And for the volume of the vacuity to be deducted, take the area of its base, computing its mean height according to its figure.

BOARD AND TIMBER MEASURE.

To ascertain the Surface of a Board or Plank. RULE.-Multiply the length by the breadth, and the product will give the surface required.

Or, lxb-surface.

NOTE.-When the piece is tapering, add the breadths of the two ends together, and take half the sum for the mean breadth.

EXAMPLE.-The length of a plank is 16 feet, and its breadth 15 inches; what is its surface?

16×1.25 (15)=20=product of length and breadth=surface required. Ex. 2. The length of a plank is 25 feet, and its breadth 14 inches; what is its surface? Ans. 29.167 square feet. Ex. 3. The length of a plank is 18 feet, and its widths at the ends are 17 and 19 inches; what is its surface? Ans. 27 square feet.

To ascertain the Contents of Squared Timber.

RULE.-Multiply the breadth by the thickness, and this product by the length, and it will give the contents required. Or, bxtxl contents.

EXAMPLE. The length of a piece of square timber is 20 feet, its sides at its less end are 15 inches, and at its greater end 19; what are its contents?

19+15÷2=17, and 172 × 20÷144=40.1388=product of square of mean side and the length÷144 to produce feet=contents required.

Ex. 2. The ends of a piece of timber are 18 and 22 inches square, and the length of it is 22.5 feet; what are its contents? Ans. 62.5 cubic feet.

NOTE.-1. If the piece tapers regularly from one end to the other, the breadth and thickness, taken in the middle, will be the mean breadth and thickness.

2. If the piece does not taper regularly, but is thicker in some places than in others, take several different dimensions, and their sum, divided by the number of them, will give the mean dimensions.