Required the solidity of a cone, the diameter of the base being 2f. 61, and the height 12f.

f. i

2 6=2.5

2.5 × 2.5 × 7854 x 12 58-905 the solidity of a cylinder of the same base and altitude.

58.905

And

= 19.635 the solidity of the cone.

3

PROBLEM III. PLATE 53.

To find the solidity of the frustum of a square pyramid. To the rectangle of the sides of the two ends add the sum of their squares; that sum being multiplied by the height, one third of the product will give the solidity.

EXAMPLE.

Let ABCDEFG be the frustum of a square pyramid, one side of the base, AB or BC, being 3f. 6i; each side, DE or EF, of the top 2f. 31; and the perpendicular height HI,

To the rectangle of the sides of the two bases add one third of the square of their difference; that sum being multiplied by the height will give the solidity.

EXAMPLE.

Let ABCDEFG be the frustum of a square pyramid, one side of the base, AB or BC, being 3f. 6i; each side, DE or EF, of the top being 2f. 31; and the perpendicular height HI, 6f. 9i; required the solidity.

To find the solidity of the frustum of a cone.

To the rectangle of the two diameters add the sum of the squares of these diameters; multiply the product by 7854, and that product by the length; then one third of the last product will give the solidity.

Note. If the circumferences be given, proceed in the same manner, only multiply by 07958 instead of '7854.

EXAMPLE.

What is the solidity of the frustum of a cone, the diameter of whose greater end is 3 feet, and that of the lesser end 2 feet, and the altitude 9 feet?

MENSURATION.

OF A PRISMOID.

123

DEFINITION. A prismoid is a solid contained under six planes, the ends being parallel, but unlike rectangles; and of the other four sides, each opposite, two are equal trapeziums. PROBLEM I. PLATE 53.

To find the solidity of a prismoid ABCDEFG.

Multiply the length at the greater end BC, by the breadth at the lesser end FE, and the length at the lesser end DE, by the breadth at the greater end AB.

To half the sum of the two products add the areas of the two ends; that sum multiplied by of the height GH, gives the solidity.

EXAMPLE.

What is the solidity of a prismoid ABCDEFG, whose greater end is 12 inches by 8, and the lesser end 8 inches by 6, and the length or height нI 5 feet?

PROBLEM II. PLATE 53.

To find the solidity of a sphere or globe.

Multiply the cube of the diameter AB by 5236, and the product is the solidity.

EXAMPLE.

What is the solidity of a globe whose diameter is 3 feet?

3x3x3=27

and 5236

To find the solidity of the segment of a globe.

To three times the square of half the diameter AB of the base, add the square of the height CD; multiply that sum by the height CD, and the product multiplied by 5236 will give the solidity.

EXAMPLE.

What is the solidity of a spherical segment, the diameter of the base being 4 feet, and the height of the segment 3 feet?

X2

2

=

2 half the diameter.

3 x 39 the square of the height.

To find the solidity of a spherical zone, the radii, ED and FB, of the two parallel circles at the ends being given, and their distance EF.

To the squares of the two radii add one third of the square

of the height; multiply the sum by the height, and the product by 1-5708 will give the solidity.

EXAMPLE.

What is the solid content of a spherical zone whose greater radius is 12 inches, and the lesser 10 inches, and the height or distance of the ends 4 inches?

122 + 102 +

42

3

× 4 × 1·5708 = 1566 6112 the solidity required.

PROBLEM V. PLATE 53.

To find the solidity of a wedge ABCDEF.

Multiply the area of the base ABCF by the perpendicular height EF, and half the product will give the solidity.

EXAMPLE.

Required the solidity of a wedge ABCDEF, the side AB being lf. 31 and BC 2f. 6i, and the height 4f.

Note. The solidity of any prismatic ungula will be found in the same manner; that is, half the product of the area of the base multiplied into the height will give the solidity.

PROBLEM VI. PLATE 53.

To find the solidity of the hoof, or ungula, from the frustum of a square pyramid.

To the square of the side of the base, or that end which is complete, add one half of the product of the sides of the two ends; this being multiplied by one third of the height will give the solidity.

If the hoofs are any other than that of a square pyramid, find the square root of the area of each end, which will give the side of a square equal in area; then proceed as above.

EXAMPLE.

Required the solidity of an ungula ABCDE, from the frustum of a square pyramid, the side of the greater end, which is complete, being 1f. 61, that of the lesser end 1f. 3i, and the height FG 5f.