CALCULATION.

First Section.

Here 60.6+ 28.7 × 15.4 = 89.3 x 15.4 = 1875.22;

and

1375.22
2

= 687.61, the area of the trapezoid ABCD.

Second Section.

Here 52.3 x 13.5 = 706.05, double the area of the triangle FGH; and 63.4 x 14.8938.32, double the 706.05938.32

area of the triangle EFH; then

1644.37

2

2

=822.185, the area of the trapezium EFGH.

Third Section.

Here 12.2 × 13.5 = 164.7, double the area of the trianangle KmN; also 13.5+ 18.3 x 27.4 31.8 x 27.4 =871.32, double the area of the trapezoid NmnM; and 23.8 x 18.3 = 435.54, double the area of the triangle 164.7871.32 +435.54 1471.56 LMn; then

2

735.78, the area of the trapezium KLMN.

Fourth Section.

=

2

Here 14.6 × 15.2221.92, double the area of the triangle SPm; also 14.6 +20.5 x 29.7 35.1 x 29.7 = 1042.47, double the area of the trapezoid SmnR; and 20.5 × 17.3 = 354.65, double the area of the triangle QRn ; 221.921042.47 + 354.65 1619.04 809.52,

then

2

the area of the trapezium PQRS.

Fifth Section.

2

Here 61.5+ 38.4 × 10.2 = 99.9 x 10.2 = 1018.98;

= 509.49, the area of the trapezoid TUVW;

and by the Rule for equidistant ordinates, we have A = 0,

38.4

B = 5.3 5.8 11.1, C 7.4, and D = = 9.6; 4

9.6
3

then A+4B2C × } D = 44.4 † 14.8 × = 59.2×

3.2 = 189.44, the area of the part VXW; then 509.49 + 189.44698.93, the area of the whole section TUVXW. To find the solid Content.

1

Here A 687.61 + 698.93 = 1386.54, B=822.185 + 809.52 = 1631.705, C = 735.78, and D = 60; then A+ 4B + 2C × } D≈ (1386.54 +6526.82 +1471.56) 60

X = 9384.92 × 20 = 187698.4 cubic feet; and 3 187698.4276951.792 cubic yards, the answer re

quired.

2. Required the number of cubic yards dug out of part of a drain or canal, from the following dimensions; each perpendicular, transverse section being divided into two triangles in the same manner as the second section EFGH, in the foregoing figure; and the common distance of the sections 100 feet:

Ans. The area of the first section

347.35; the areaof the second 341.735; the area of the third = 380.13; the area of the fourth = 379.55; the area of the fifth = 394.505; the area of the sixth = 393.1; the area of the seventh =403.63; and the content = 2252593 cubic feet 8342 cubic yards, 25 feet, the answer required.

MARLPITS.

MARL is a kind of rich clay, and is used as manure for land, in Lancashire, Cheshire, Derbyshire, and other counties in England; and is commonly dug by the cubic yard.

Marlpits are of various forms. Sometimes they are laid out in the shape of a rectangle, sometimes in that of a trapezium, and sometimes as an irregular polygon.

In digging a marlpit the sides are always sloped, in order to prevent the upper edges from slipping in; and as the bottom is seldom perfectly level, there is generally a little variation in the depths.

PROBLEM IV.

To find how many cubic yards have been dug out of a marl-pit.

CASE I.

When the top and bottom of a marl-pit are rectangles, the pit may be considered as a prismoid.

RULE.

To the sum of the areas of the top and bottom, add four times the area of a section half-way between them; multiply this sum by the mean perpendicular depth, and of the product will be the solidity.

Note 1. When the sides and ends are regularly sloped, the length of the middle section will be equal to half the sum of the lengths of the top and bottom, and its breadth equal to half the sum of their breadths; but if the inclination of the sides and ends be not regular, the length and breadth of the middle section must be found by actual admeasurement.

2. A mean depth must be found by taking several depths, at equal distances from each other, and dividing their sum by their number.

EXAMPLES.

1. Let the subsequent figure represent a marlpit, the dimensions of which are as follow: viz. the length AB of the top = 58.6 feet, and the breadth BC = 36.2 feet; the length EF of the middle section= 55.7 feet, and its breadth FG 33.4 feet; the length KL of the bottom 52.8 feet, and breadth LM = 30.4 feet; how many cubic yards of marl were dug out of the pit, its mean perpendicular depth being 8.6 feet?

Here 58.6 × 36.2=2121.32, the area of the top; 55.7 X 33.4 X 4 = 1860.38 x 4 = 7441.52, four times the area of the middle section; and 52.8 × 30.4 — 1605.12, the area of the bottom; then (2121.32 + 7441.52 + 1605.12) x 8.6 = 11167.96 × 8.6 = 96044.456, which being divided by 6, we obtain 16007.409 cubic feet = 592.867 cubic yards, the answer required.

2. A marlpit measures 86.4 feet in length, and 36.8 feet in breadth at the top; the length of the middle