MEN

188

MEN

i

The same found by the rule thus :

5348

5

3)26740

__ R. Qr. Ft. In.

272)8913 4 (32 3 5 4, the answer, as816 before.

753

544

68)209(3

204

5

The rule here applied is the same as Rule I. p. 50, vol. I.As the first progression goes on from 1 to 50 inclusive, andtwo adjacent numbers differing by unity, any number from50 to 1 may be taken at once.

Example. —In a wall of two bricks thick, containing anarea of 47 feet, what is the standard measure ?

In the meeting of the 2-brick column and the horizontalrow from 47, you have 62 feet 8 inches, the answerrequired.

In constructing a table, such as this, with differentdenominations, it will be much more ready to find the wholenumbers, as in the last Table, in the least denomination, andafterwards to divide each respective number by as many aswill make one of the next denomination ; then the quotientis so many of this denomination, and the remainder, if any,so many of the first; and if the quotient exceed the numberthat makes one of the third denomination, divide it by thisnumber, and the second quotient is so many of the thirddenomination, and the remainder, if any, so many of thesecond ; and so on, as long as division can be made.

Table IV.

Explanation and Use. —As the prices of bricks and oflabour may vary in any ratio, in respect of each other, it isconvenient to have a price at once agreeable to these varia-tions, which can be easily obtained from this Table. Thegiven price of bricks per thousand is arranged in the firstvertical column, the price of mortar and labour in the givenhorizontal row, and the required price per rod is found, asin the other Tables, in the point or place of intersection ofthe horizontal row and vertical column, carried from eachof the given prices.

Example. —What is the price of a rod of brickwork, whenthe rate of bricks is £1 18s. per thousand, and the price ofmortar and labour £3 10s. per rod ?

Carry your eye horizontally from £1 18s. in the givencolumn of bricks, until you come to the vertical columnunder £3 10s. the given price of labour and mortar ; and youwill find £12 Is. the price of the rod, as required.

Construction. —Multiply the price of bricks per thousandby 4lj, and add the product to the price of labour at the top,and you will have the price per rod, as in the Table. Letus take the above as an example : then (£1 18s.) x 4^ =£8 11s. the price of bricks in a rod, to the product; add£3 10s. the price of labour and mortar, and you will have£12 Is, as before.

This Table might be otherwise constructed, by giving aTable of the variations of lime and sand, but then it wouldbe much longer. The articles of labour, lime, and sand, areclassed together, upon the supposition or probability, thatthe increase or decrease of either of the two is nearly in thesame ratio ; but should it prove otherwise, and the dis-

parity be great, a Table must be constructed upon dineprinciples.

Table V. t(f o

Explanation and Use .—This Table eonsists of only .vertical columns ; the price per foot is contained in theand the price per rod in the second column. So that »price per foot be given, you have the price per r0 .^ 9the same horizontal line ; or the very reverse is somet'necessary, when the price per rod is given, to knowprice per foot.

Example .—What will 3r. 258ft. of reduced brick ^come to, at £2 16s. 8d. per rod ? First multiply £2 1° 3 'by 3 ; look for £2 16s. 8d. under the column per rod, 00 g ,left hand of it, you have O^d. for the price per foot; * „

fore multiply 258 feet by (r, that is, divide it by 2, and 7.jjwill have the number of pence, which divided by 1*give the shillings. Thus :

£•

2

16

d.

8

3

10

10

Feet.

2)258

12)129

10 9

9 0 9

To perform such operations entirely by rule is 30111very tedious. . a

Construction .—The quantity of superficial feet in . fe sis 272 ; this being multiplied by the price per {ooU jjevidently the price per rod : Suppose Ofd. per foot, Wthe priee per rod ?

Thus: 272

multiplied by 3

4)816

12)204 pence per rod.

17 shillings.

Table VI. . , [n tb e

Explanation. —The different gauges are containe jpotop row, the given areas in the first vertical cointhe answers in the succeeding vertical columns.

Example. —How many tiles are sufficient, at agauge, to cover an area of 5,349 feet ?

Here 5349 = 5000 + 300 + 40 + 9.

Now, 5000 will require 32,500300 „ „ 1,950

J9 ;; ;; Jk

therefore 5349 “ „ 34,768^ nU ujh^

Or thus, by calculation :—It is evident that the ^ t j]are as the areas : suppose it is ascertained that o&will cover 8 feet superficial, at a 7-inch gauge, S £ ee t, s ° J

have, as the area of 8 feet, is to the area of 5,349 t, e r8 tiles, the number required for 8 feet, to Iketiles required, 34,768^.

Thus 8 : 5349 : : 5252

10698

26745

8)278148

34768£ the answer, as