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Where the foundation would consist of several straightlines, forming trapezoids, the best method is to find the con-tent of each trapezoid separately ; and then adding all thetrapezoids together, their sum will be the area of the whole ;but if the figure of the ground on which the building israised be a curve, the measurer will be grossly deceived asto the true contents of the work, unless he divide the lengthinto equal parts, as already recommended.

The contents in brickwork may be found by multiplyingthe area by 3, and dividing by the number of half bricks.

The following example is added, in order to show the useof the method of equidistant ordinates.

Example. —Let e f g h be a wall of brickwork, or theback of a house, to be built over a public road, or valley,hls; the under part of the wall is built from the founda-tion, H l G, up to the level at I K, three brieks thick, andfrom ik to the top, e f, parallel to it, two bricks and a halfthick, to the height of 15 feet, having five windows in it;the vacuities on the outside of each window are 8 feet by 4feet, and half a brick thick ; the vacuities on the inside are8 feet by 4 feet 9 inches, two bricks thick; the recess onthe inside for the finishing of the backs in each window, oneand a half brick thick ; the height 2 feet 6 inches, from thetop of the floor to the sill of the window ; the width is thatof the vacuity on the inside of the window, viz., 4 feet 9inches. There is an arched way underneath for carriages,&c. to pass through, whose opening is 12 feet, and its heightfrom the level of the pavement to the crown or top of thearch 11 • feet, and the height, from the pavement to thespringing of the arch, 9 feet; the under wall is divided intoan even number of equidistant spaces, whose ordinates arerespectively as follow :—6 feet, 10 feet, 13 feet, 14 feet, 10feet, 4 feet 6 inches, and 1 foot; the whole length of thebuilding is 50 feet; required the number of bricks, and thequantity of sand and lime to build the said wall.

Explanation. —The under part of the building being anirregular figure, it is measured according to the method ofequidistant ordinates, Problem VI.; the upper part is foundas in the foregoing examples. The arched way is measuredby Problem V. The contents of the windows are obtainedby Problem I. Then deduct all the vacuities at half a brickthick from the area of the whole, found as if it were solid, athalf a brick thick, as before : the remainder being divided by3, will reduce it to the standard thickness of one brick and ahalf.

Feet.

In.

Feet.

10

0

13

14

0

10

4

6

—

23

28

6

2

4

46

114 0 four times the sum of the even ordinates,

add 46 twice the sum of the odd ordinates.

3 ) 160

53 450

6 ) 2666 8 [thick.

444 5 area of the under part of the wall, 3 bricks6 number of half bricks.

2666 8 area of the under part of the wall, £ a brick•-" thick.

Feet. In.

15 0 height of the upper part of the wall-50 0 length.

750 0 area of the upper part of the wall, 2£5 0 number of half bricks.

- brio * 9

3750 0 area of ditto, a brick thick.

8 04 0

[ tb ick.

, bfl c *

32 0 area of the vacuity on the outside, i 9

5 0 number of windows.

' e fit®

160 0 area of the vacuities on the outside o

- windows, ^ a brick thick.

8 44 9

33 46 3

- [window, 2 bricky 0 p 0

39 7 area of the vacuity of the inside

5 number of windows. r .„ aid®'

Kv

197 11 area of the vacuities for five windows4 half bricks thick.

- . . wbrf®*

791 8 area of the vacuities on the inside, %

- thick.

4 92 6

9 6

2 4 6 .. t thi c,c '

- [each window 1J l* 11

11 10 6 area of the vacuity of the reces

- [backs, !***>*

59 4 6 area of the vacuities of the fi ve

3 number of half bricks.

178 1 6 area of the vacuities of the

w> 9 '

do*

- backs, -Jj a brick thick.

11 0 height of the archway, from the P aV

2 the crown.

e0 {

22 0

add 9 0 height from the pavement to the 3 P r

•- the arch.

31 0

10 O width of the archway.

ing

iog

3 ) 310 0

103 4 area of the vacuity of the archw®/'

6 number of J bricks thick.

620 0 area of the vacuity

- thick.

160 0791 8178 I620 0

the vacuity of the archwa/’ £

1749 9 areas of all the vacuities, £ 9