BRI
'54
BRI
from the whole; the remainder is divided by 3, which givesthe superficial contents in feet of the surface of a wall reducedto 1 % brick in thickness; the deductions being negative quan-tities, no farther notice is taken of them; the sums of theother columns being positive, the price is affixed to eachcommon measure, whether a foot or a rod, &c. and the valueof each quantity is found by this common measure; then thequantities, with the prices of their common measures andvalues, are inserted in a bill; the whole being reduced intoa sum, gives the amount of the whole money for the wall.
Abstract.
Footing of wallof place-bricks.
The whole of the walling re-duced to £ a brick thick, takenas gray-stock work withoutregard to the facing.
Contents of the wall.
Deductions.
3)221 0
1716 0
1430 0
48 0
291 8
73 8
1352 0
68 0
780 0
163 4
68 0
5278 0
81 8
2103 3
301 0
68 0
3)3174 9
114 9
12 6
272)1058(3 R.
816
242
23 10120 0430 6
81 098 0
133 0
2103 3
Quoins ofapertures bythe linealfoot.
73 051 6112 051 0183 0124 6
595 0
Rubbedand gaugedarches.
Marl facing in superficial feet.
Contents.
Deductions.
ft. in.
ft. in.
ft. in.
37 1
884 0
68 0
16 6
309 3
120 0
81 0
CO
0
57 9
34 0
6 3
309 3
The next thing to be done is to affix the price of thecommon measure to each of the above species of work, andfrom this, to calculate the quantity of each ; then insert theseveral sums in a bill, as follows :
Bill.
Rods. Feet. £. s.
0 73§* Superficial of place-brick work reduced 1 „ „to the standard at £12. 15s. per rod j3 242 Superficial of gray-stock work reduced to ) a
the standard at £14. 10s. per rod.J 00 °
0 595 Feet lineal of quoins in apertures at Id 1 „ a
per foot. . S
0 531 Superficial of rubbed and gauged work ) „
at 3s. per foot . 5
0 574J Superficial of marl facing at 6d per foot 14 7
£ 84 13 10}
* f not noticed in the calculation.
In the foregoing admeasurement, the quoins are valuedby the foot lineal, in addition, and the method of making outthe contents is different from that commonly used ; butthough the allowance is strictly just, and the practice shorterand less liable to mistake than that in common use; that
d.
H
0
7
6
45
there should be nothing wanting to gratify those who maystill favour the customary method, the admeasurement ishere repeated in the usual way. The mark thus — signifiesa deduction.
Dimen
Contents.
Th.
Names.
ft.
26
1
in.
0
0
ft. in.
261 0
brick
4 1
Footing.
26
11
0
0
286 0
3
Walling
2
6
4
0
0
— 48 0
i
Windows
2
6
4
3
8
— 58 4
n
Ditto
8
4
6
0
— 34 0
1
Doorway
8
4
9
8
— 40 10
2
Ditto
26
11
0
0
286 0
25
Walling
2
8
4
6
0
— 68 0
4
Windows
2
8
4
9
8
— 81 8
4
Ditto
2
10
4
9
8
—100 4
U
Ditto
8
4
6
0
— 34 0
1
Ditto O
8
4
6
6
— 38 3
4
Ditto O
4
2
0
0
— 63
1
Ditto
4
2
6
3
— 7 11
H
Ditto
26
13
0
0
338 0
2
Walling
3
10
4
0
0
—120 0
i
Windows
3
10
4
3
8
—143 6
14
Ditto
26
10
0
0
260 0
ii
Walling
3
6
4
9
0
— 81 0
4
Windows
3
7
4
0
8
— 98 0
4
Ditto
3
9
4
6
8
—133 0
4
Ditto
The dimensions are frequentlysquared upon the waste on themargin of the Dimension Book $the following is a specimen:
First.4 8
6
3
Second.
8 9
4 8
28 0
35 0
1 2
29 2
2 11
2 11
2
40 10
58 4
Third.
Fourth.
8 9
4 8
4
8
8 9
5 10
0
3 6 0
35 0
37 4
40 10
0
40 10 0
The operations are wrought invarious ways: as in the first andsecond, by aliquot parts; the thirdand fourth by duodecimals. These
various modes serve to confirm
each other. The third and fourthare the same dimension, but thefactors are inverted in respect ofeach other.
4 4 6
2 2 3
0 8 1169 5 9 0
0 3 4 10 1 6
6 0 0 9 5
6 3 4 4 2 7 6
- 7 7 16
7 11 4 1 6
The mark thus O signifies asemicircle. The method of find-ing the semicircular area is shownin the two examples above; firstby multiplying the dimensionstogether, and multiplying the pro-duct by 9 inches and 5 seconds, asabove, or by calling the feet inches,as in the first operation, and multi-plying by 9 feet 5 inches.
The above contents are abstracted by inserting each areain a separate column, according to the number of half bricksit is thick.
In the following abstract, the several parts of each columnare collected : the sum or amount is multiplied by the numberof half bricks, and the product divided by 3 ; then the positivequantities are added together, and the negative ones addedtogether, their difference is the real quantity in reduced feet,by dividing which by 272, the answer will be obtained in rods.