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(PART V.) Now, 2.7652978 x 40 138.26489 feet, the content of the solid part, which being taken from the whole upright space will leave the content of the vacuity within the

room.

2. What is the solid content of a saloon with a circular quadrantal arch of 2 feet radius, springing over a rectangular room of 20 feet long, and 16 feet broad? Ans. 580.2065 cubic feet.

3. A circular building of 40 feet diameter, and 25 feet high to the ceiling, is covered with a saloon, the circular quadrantal arch of which is 5 feet radius ; required the capacity of the room in cubic feet?

Ans. 30766.496 cubic feet.

PROBLEM VI.

To find the superficial content of a saloon.

RULE.

Find its breadth by applying a line close to it across the surface; and its length by measuring along the middle of it, quite round the room; then the product of these two dimensions will be the surface required.

Note. The area of the flat ceiling must be added to the area found by the above Rule, in order to obtain the whole surface of the saloon.

EXAMPLES.

1. The girt across the face of a saloon is 5 feet 3 inches, and its mean compass 94 feet 6 inches; what is the area of its surface?

Here 94.5 × 5.25 = 496.125 feet, the area required. 2. The mean compass of a saloon is 126 feet 10 inches, and the girt across its face 8 feet 6 inches; what is the area of its surface? Ans. 1078 ft. 1 in.

PROBLEM VII.

To find the solid content of the vacuity formed by a groin arch, either circular or elliptical.

RULE.

Multiply the area of the base by the height, and the product by .904, and it will give the solidity required.

Note. Groins are sometimes measured as if they were solid, in consideration of the great trouble and waste of materials in forming the arches and intersections.

EXAMPLES.

1. What is the content of the vacuity formed by a circular groin, springing from the sides of a square base, each side of which is 14 feet?

2. What is the solid content of the vacuity formed by an elliptical groin; the side of its square base being 24 feet 6 inches, and its height 8 feet 3 inches ?

Ans. 4476.6645 feet.

PROBLEM VIII.

To find the concave surface of a circular or an ellip

tical groin.
RULE.

Multiply the area of the base by 1.1416, and the product will be the superficies required.

Note. In measuring works where there are many groins in a range, the cylin drical pieces between the groins, and on their sides, must be taken separately.

EXAMPLES.

1. What is the concave surface of a circular groin arch; the side of its square base being 15 feet 6 inches? Here 15.5 x 15.5=240.25, area of the base; and 240.25 × 1,1416 274.2694 feet, the answer required. 2. The base of a groin is a rectangle whose sides are 20 and 26 feet; required the concave surface of the arch. Ans. 593.632 feet.

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GENERAL ILLUSTRATION.

HAVING gone through the Works of Artificers, and noted the methods of measuring buildings, and computing their contents; I now proceed to give a general illustration of the whole, by assigning the dimensions of a house, and from thence computing the contents of the works of the different Artificers employed in building it.

In performing this task, are shewn the methods of ruling the book, entering the dimensions, with the contents; then the method of abstracting the contents; and lastly, of forming the bills of expenses of the work and materials.

The building of which I have made choice, for the general illustration, consists of two stories, beside the cellars, and of two rooms upon each floor; which will be found quite sufficient to exemplify the methods of measuring the works of Artificers.

The whole length of the building, on the outside, is 53 feet 6 inches, and its breadth 24 feet. The walls of the cellars are 6 half bricks, or 24 inches in thickness; those of the lower story 20 inches; and those of the upper story 16 inches.

One of the lower rooms is considered to be the kitchen, and the other the parlour; and they are frequently distinguished by these denominations, in the following notes.

The plans of the different stories, and the elevation, could not be given without a folding plate; but it is presumed that the reader will find no difficulty in comprehending the dimensions without them.

Note. The columns of numbers, in the following forms, are sufficiently explained by the titles at the tops of them; excepting the figures 2, 3, &c. in the first column; which figures signify that there are more than one article of the same dimensions: consequently the contents arising from the dimensions to which these figures are prefixed, must be multiplied by 2, 3, &c.; and the products entered in the column of contents.

2

3

Half

Dimensions Bricks CONTENTS.

thick.

TITLES.

Ft. In.

Ft. In.

247 O

Chimney shafts of the ground story.

204 2 Ditto of the upper story.

90 0 Ditto of the gable ends. The deductions are as follow; viz.

47 10 Cellar doors.

4

3

93 4 Doors in partition walls.

In order to abstract the foregoing contents; that is, to collect them into one sum, make the deductions, and reduce the neat contents to the standard thickness of one brick and a half, proceed thus: Make only two columns for the whole contents, and two for the deductions of the same thickness; viz. one column for the contents that are one brick in thickness, and the other for the contents that are one brick and a half in thick