(20) The area of the base of a parallelepiped is 124ft. 28in. and its altitude is 8ft. 4in.: find its content in feet and inches.

Answer: 1034ft. 1648in.

(21) A parallelepiped contains 94ft. 235in. and the area of its base is 24ft. 5in.: find its altitude.

Answer: 3ft. 11in.

(22) Find the numbers of feet and inches in the side of a square whose area is 1122ft. 36in.

Answer: 33ft. 6in..

(23) Find the edge of a cube which contains 15 solid feet and 1080 solid inches.

Answer: 2ft. 6in.

(24) The area of one side of a cube is 12. 3'; find its capacity.

Answer: 42 cubic ft. 1512 cubic in.

(25) Divide 1532. 9'. 9" superficial measure by 81. 9' lineal measure.

Answer: 18ft. 9in.

(26) How much in length that is 1ft. 2in. broad and 1 in. thick, will make a solid foot?

Answer: 6ft. 10 in.

(27) A gardener has a piece of matting 73yds. 1f. 8in. long and 3ft. 9in. broad, to cover a wall 94ft. long and 10ft. high how much of the wall will be left uncovered?

Answer: 112 ft.

(28) On laying down a plot of ground with sods 2ft. 6in. long and 9in. wide, it is found that it requires 75 sods to form one strip extending its whole length, and that a man can lay down 11 strips each day: find the surface covered in 8 days.

Answer: 14061ft.

(29) The roller used for a bowling-green being 6ft. 6in. in circumference and 2ft. 3in. in width, is observed to make 12 revolutions from one extremity of

the green to the other: find the area rolled, when the roller has passed 10 times the length of it.

Answer: 195yds.

(30) A reservoir is 24ft. 8in. long and 12ft. gin. wide: how many cubic feet of water must be drawn off to make the surface sink 1 foot?

Answer: 314ft.'

(31) If a cubical block of stone contain 14706ft. 216in.: find the length of all its edges, and the area of all its faces.

Answers: 294ft. and 36014ft.

THE COMPUTATIONS OF GAGERS.

214. DEF. The dimensions made use of by Gagers are taken in inches and parts of an inch expressed decimally and from them, the contents of cisterns, malt-bins, &c., are computed by such rules as have been already laid down, and they will therefore be expressed in cubic inches and their decimal parts.

215. Liquids are always estimated by the imperial Gallon which is equal to 277.274 cubic inches: and therefore, when the content of a vessel has been ascertained in cubic inches, the number of gallons it contains will be found by dividing it by 277.274.

Ex.

What number of gallons are contained in a cistern whose length is 40 inches, breadth 24 inches and depth 16 inches?

=55.3964 &c. gals. = 55gals. 1qt. 1pt., nearly.

216. Malt, Corn, &c., are always estimated by the imperial Bushel consisting of 2218.192 cubic inches; and the number of bushels will be obtained by dividing the content, ascertained as before, by 2218.192.

Ex. If a circular room 5 feet in radius, be filled with malt to the depth of 6 inches: find the number of bushels it contains.

Here, the content = 3.14159 × 60 × 6

= 67858.344 cubic inches:

=

67858.344

and the number of bushels = 2218.192

= 30.5917 &c. bush. 30 bush. 2 pks., nearly.

=

Whenever the depth is an inch, the content of any upright vessel or cistern is expressed by the area of its surface and in this sense the term surface is used in gaging.

Examples for Practice.

(1) How many gallons are contained in a cubic foot? Answer: 6.232 gallons, nearly.

(2) The length of a cistern is 169 inches and the breadth is 125 inches: how many gallons does it contain, the liquor being 4 inches deep?

Answer: 304.752 gallons, nearly.

(3) If the interior edge of a cubical vessel be 1 ft. 3in., how many imperial gallons will it hold?

Answer: 12.172 gallons, nearly.

(4) What is the content of a cylindrical vessel, the radius of whose base is 20 inches and height 54 inches? Answer: 244.733 gallons, nearly.

(5) What number of bushels are contained in the space of a cubic yard?

Answer: 21.033 bushels, nearly.

(6) How many bushels of malt are there on a floor 5 feet by 4 feet, when its depth is 14 inches?

Answer: 20 bushels, nearly.

(7) If the exterior dimensions of a closed rectangular bin be 3ft. 5in., 2ft. 4in. and 1ft. 3in., find the quantity of malt it will contain, if the thickness of the material of the bin be lin.

Answer: 6.816 bushels, nearly.

(8) The diameter of the base of a standard bushel measure is 18.789 inches: find its height.

Answer: 8 inches, nearly.

The practical calculations of Excisemen are greatly facilitated by means of an instrument called a Sliding Rule and by Tables containing the proper multipliers

H.A.

10

and divisors for Squares, Circles, &c., which may be seen in any work treating expressly upon the subject.

THE COMPUTATIONS OF LAND-SURVEYORS. 218. DEF. The dimensions of land or of any surface of considerable extent, are taken by means of Gunter's Chain which is 4 poles or 22 yards in length and is divided into 100 equal parts called Links.

219. Since an acre is equal to a parallelogram, 40 poles or 10 chains in length and 4 poles or 1 chain in breadth, it will contain 1000 x 100 = 100000 square links; and therefore, if the lineal dimensions be expressed in links and the superficial contents be found, these results when divided by 100000 or with five figures cut off towards the right hand, will be the numbers of acres.

A lineal pole being 5 yards or 25 links, the magnitude of a square pole will be

5×5 = 30 sq. yards; or, 25 × 25 = 625 sq. links: so that a pole = 625 sq. links or = 30.25 sq. yds.: a rood = 25000 sq. links or = 1210 sq. yds.: an acre = 100000 sq. links or = 4840 sq. yds. Hence also, the magnitude of a square mile

= 1760 × 1760 = 3097600 sq. yds. = 640 acres.

Ex. The length of a rectangular field being 25 chains 8 links and its breadth 14 chains 75 links: what number of acres does it contain?

and therefore the field comprises 36 ac. 3 ro. 3833 po.

Hence also, if the length of a field containing 36 ac. 3 ro. 383 po. be 25 chains 8 links, its breadth will be found by reversing the operation upon these magnitudes when expressed in links thus,

:

1475 links = 14 chains 75 links.

Examples for Practice.

(1) Find the area of a square field whose side is 10 chains.

Answer: 11 ac. 4 po.

(2) The base of a triangular field is 16 chains 3 poles and its perpendicular is 6 chains 2 poles: what number of acres does it contain?

Answer: 5 ac. 1 ro. 31 po.

(3) The sides of a triangular field are 380, 420 and 765 yards: how many acres are contained in it? Answer: 9ac. 38 po., nearly.

(4) The diagonal of a trapezium is 5 chains and the perpendiculars upon it from the opposite angles are 3 and 2 chains: find its area.

Answer: 1 ac. 1 ro. 31 po.

(5) A field in the form of an ellipse has its greatest and least diameters equal to 7 and 5 chains: find how many acres there are in each of the parts into which they divide it.

Answer: 2 ro. 30 po., nearly.

(6) One acre of land is to be cut from a rectangular field whose breadth is 2 chains 50 links, by a line parallel to it: find the length of the plot.

Answer: 4 chains.

(7) What is the length of the side of a square field comprising 2 acres 4 poles?

Answer: 4 chains.