MEN
181
MEN
t H-—TV: measure any angle chimney, standingy distant each way from the angle of the room.the tlpl y *be breadth, a b, by the height of the story, andh a |p p , ro< ^ UK t by the number of half bricks contained in thethick rea< ^*h’ A B > and it will give the solidity at half a brickehi m ’ after deducting the vacuity, or opening of the
p ie y>
plane LH -—To measure an angle chimney, when the
e 9uall ltS ^ reas t intersects the two sides of the room un-j. * distant from the angle.
ititerg 001 ^ le P°ints A and b, where the plane of the breastPar a ii , the sides of the room, draw two lines, a e, e b,of to the two sides of the room; then multiply eitherr° 0m nnes, a e or e b, suppose e b, by the height of the
. i » ’ ri • * * o
oriek, aa Multiply that product by the number of halftts bef C ° n t a ined in the other line, ae, and deduct the vacuity4 brigand the remainder will be the content, at halfj, ^ thick.
IV .—To measure an angle chimney, when thedistant breast projects out from each wall, and. unequally
Hra d r ° m the angle of the room.the ro t ie two lines, g p and r h, parallel to the two sides of^ as before ; then multiply the breadth, p h, by the°f t}, e ot l be story, and the product contained in the halfPli e( j, ot ber side, fg; from this product deduct pd, multi-^ * .height of the story, and hy the number of halfthe i . c °ntained in the half of P C, and also the vacuity of£ C, nniney.
• To°t Bl *- EM ^•~To find the area of an arched aperture.l ar »bs • WlC f t^ le height at the middle add the height of the
tl.,
:a I for
and one-third of the sum multiplied by the breadthaperture will give the superficial content, sufficiently
5 »—Let the height of the arch be 12 feet, eachthe eet - and the breadth of the aperture 5 feet; what8up erficial content ?
Feet.
12
2
24
10
3 ) 34
"iii
5
56$ feet, the answer.
tb '“‘8 fro ^ reater accuracy be required, add the quotientft 6 breadth* t f' e < ^ v * s i°n of the cube of the altitude by twice
^ the truth ^ aperture > an< I tbe sum will be exceedingly
'
'sit if
, the height of the
“m —In the foregoing examp ’ twice the breadth
of tv' 8 2 feet, and the chord of the ar > of 2 is 8 ; and 8
^idT rtWe ’ is 5 fe6t 5 the -“ 8 of a foot for the quan-tiA 10, or twice five, gives .8 ot a mu
v 0 be added to the above-^ the above 56$ = 56-666, &c.
-8 [the truth.
STA66, the area-
'Von 141 ;' 1 '! VI.— To find the area of a ™ a ^ %c°iwpP 0Singit f 0 hZ a building, placed upon a curve JVo. built upon uneven ground.
^ ; TI - 3a
Divide the length into any number of even parts (themore the truer) by parallel vertical sections ; call the first ofthese sections odd, the next even, the next odd, and so onalternately ; and thus the first and last sections will alwaysbe odd. Then add together four times the sum of the oddsections, twice the sum of the even sections, and the twoends: divide the sum by three times the number of partsthat the length is divided into ; then multiply the quotientby the length of the wall, and the product will be the super-ficial content.
Example .—Suppose a brick wall, 32 feet long, to bedivided into eight equal parts, by seven sections of the fol-lowing heights, taken in successive order, 5 feet, 6 feet, 6feet 2 inches, 6 feet 3 inches, 5 feet 5 inches, 4 feet, 3 feet,the one end 3 feet, and the other 2 feet 6 inches; requiredthe area of the wall.
Now, 5 feet, 6 feet 2 inches, 5 feet 5 inches, and 3 feet,are the odd ordinates ; and 6 feet, 6 feet 3 inches, 4 feet, arethe even ordinates ; therefore,
78 4 = four times the sum of the odd sections.32 6 = twice the sum of the even sections.
3 0 = the one end.
2 6 = the other end.
3)116 4
8 ) 38 9 4
4 10 232
155 1 4
The following erroneous method is usually practised bymeasurers and workmen :—
Add all the heights together, and divide the sum by theirnumber ; then the quotient, multiplied by the length, issupposed to give the area.
Let us therefore resume the same example by this method:
3
0
5
0
6
0
6
2
6
3
5
5
4
0
3
0
2
6
)41
4 the i
4
7
i*
■ 32
146
ii
6}
the truth.