PROBLEM II.

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?

3 X 3 X 3 = 27

And ⚫5236

27

36652 10472

14 1372 the solidity of the globe.

PROBLEM III.

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 the sum by the height CD, then 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 fect, and the height of the segment 3 feet?

12 three times the square of half the diameter.

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

To the squares of the two radiusses, 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 X 4 X 1.5708 1566-6112 the solidity required. 3

PROBLEM V.

To find the solidity of a wedge ABCDE.

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

EXAMPLE.

Required the solidity of a wedge ABCDE, the side AB being İf. 3in. and BC 2f. 6in. and the height 4f.?

2.5 x 1.25 × 4

Then

6'25 the solidity.

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.

VOL. I.

D d

PROB

PROBLEM VI.

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, gives the solidity.

And 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. 6in. that of the lesser end 1f. Sin, and the height FG 5f.?

To find the solidity of a spheroid; the fixt axis and the revolving axis being given.

Multiply the fixt axis by the square of the revolving axis, and the product by 5236, will give the solidity.

EXAMPLE I.

What is the solidity of a prolate spheroid, whose transverse aris is 100 feet, and the conjugate 60 feet?

100 x 602 x 5236 188496 the solidity required.

EXAMPLE II.

What is the solidity of an oblate spheroid, whose transverse aris is 100 feet, and the shortest axis 60 feet?

60 × 1002 × 5236314160 the solidity required. .

PROBLEM VIII.

To find the solidity of a parabolic conoid; the diameter AB of the base being given, and the perpendicular height CD.

Multiply the square of the diameter of the base by 3927, and the product by the height will give the solidity.

EXAMPLE.

What is the solidity of a parabolic conoid, whose height is 50 feet, and the diameter of the base 30 feet?

302 × 3927 × 50 176715 the solidity required.