breadth BC 8 inches; the length EF of the less end 12, and its breadth FG 6 inches; and the perpendicular height PE 18 inches?

Here 18 x 8+ 12 × 6= 144 + 72 216, the sum of the areas of the two ends.

dle section; then 15 X 7 X 4 = 105 X4= 420, four times

18

the area of the middle section; whence 216 + 420 × 6 =636 × 3 = 1908 cubic inches, the solidity required.

2. How many solid feet of timber are contained in a beam whose ends are rectangles; the length and breadth of the greater being 32 and 20 inches; the length and breadth of the less 16 and 8 inches; and the perpendicular length 25 feet? Ans. 61 feet.

3. The perpendicular altitude of a stone pillar is 8 feet 6 inches; the length and breadth of the greater end are 26 and 16 inches; and the length and breadth of the less end 18 and 10 inches; required the solidity of the stone. Ans. 17.118055 feet.

4. The length and breadth of a fish-pond at the top, are 132 and 64 yards; the length and breadth at the bottom 116 and 48 yards; and the perpendicular depth 12 feet 9 inches; what did it cost digging, at 63d. per cubic yard? Ans. £832 11s. 6d.

5. What is the capacity of a coal-waggon, the inside dimensions of which are as follow: at the top, the length is 80, and breadth 56 inches; at the bottom, the length is 41, and the breadth 32 inches; and the perpendicular depth 47 inches? Ans. 129814 cubic inches.

Note. Coals are frequently carried from pits in prismodial waggons, having four cast-iron wheels, which run upon a rail-way

They are prevented from leaving the rail-way by a particular construction of the wheels; the inner edge of the rim of each wheel being made to project. Coals are conveyed, in waggons of the above description, from the pits in the vicinity of Newcastle, to the river Tyne; from those near Sunderland, to the river Wear; and from Middleton Colliery, to the town of Leeds.

In the last question are given the dimensions of a Leeds coal waggon, which differs but little from those of other places.

Instead of rails, the waggon-road between Middleton and Leeds, is made with bars of cast-iron; and the waggons are drawn by a machine which moves along the road before them, by the force of steam. This machine was invented by Mr. Blenkinsop, of Middleton; and is called "Blenkinsop's Patent Steam Carriage." It is considered to be one of the most curious pieces of mechanism that ever appeared in England, or perhaps in the world.

It moves at the rate of about three miles and a half in an hour; and is capable of drawing 30 waggons at once, upon level ground, which are computed to weigh 105 tuns.

PROBLEM XI.

To find the solidity of a sphere or globe.

RULE.

Multiply the cube of the diameter by .5236, or the cube of the circumference by .016887, and the product will be the solidity.

Note 1. The convex surface of a sphere may be found by multiplying the circumference by the diameter. Or multiply the square of the diameter by 3.1416, and the product will be the convex surface.

2. The solidity of a sphere is equal to two-thirds of the solidity of its circumscribing cylinder.

3. The surface of a sphere is equal to 4 times the area of a circle of the same diameter as the sphere; or to the area of a circle whose diameter is double that of the sphere; or to the convex surface of the circumscribing cylinder.

EXAMPLES.

1. What is the solidity of a globe whose diameter AB is 25 inches?

B

Here 253 × .5236=25 × 25 × 25 × .5236 = 625 × 25 × .5236 = 15625 × .5236 8181.25 cubic inches, the solidity required.

2. What is the solidity of a spherical stone whose circumference measures 5 feet 6 inches?

Ans. 2.80957 feet. 3. If the circumference of a cannon-ball be 18.6 inches; what is its solidity? Ans. 108.66541 inches. 4. The diameter of the moon is 2180 miles; what is her solidity in cubic miles ?

Ans. 5424617475.2 miles. 5. The diameter of the earth is 7957 miles; what is its solidity in cubic miles, and its convex surface in square miles ?

Ans. The solidity is 263858149120 cubic miles, and the convex surface is 198944286 square miles.

PROBLEM XII.

To find the solidity of the segment of a sphere.

RULE I.

To three times the square of the radius of the segnient's base, add the square of its height; and this sum multiplied by the height, and the product again by .5236, will give the solidity.

RULE II.

From three times the diameter of the sphere subtract twice the height of the segment; multiply the remainder by the square of the height, and that product by .5236 for the solidity,

Note 1. The surface of the segment of a sphere may be found thus: Multiply the circumference of the whole sphere by the height of the segment, and the product will be the convex surface; to which add the area of the base, and the sum will be the whole surface.

2. The area of a spherical triangle, or the spherical surface included by the arcs of three great circles of the sphere, intersecting each other, may be found by the following rule:

As two right angles, or 180 degrees,

Are to the area of a great circle of the sphere;

So is the excess of the three angles above two right angles,
To the area of the triangle.

EXAMPLES.

1. The radius An of the base of the segment ABC is 9 inches, and the height Cn 7 inches; what is the solidity of the segment?

n

=

By Rule 1., we have 92 × 3 + 72 = 81 × 3 +49 24349292; and 292 x 7 x .5236 = 2044 × .52361070.2384 inches, the solidity required.

2. If the diameter of a sphere be 3 feet 6 inches; what is the solidity of a segment whose height is 1 foot 3 inches? Ans. 6.545 feet. 3. What is the solidity of a spherical segment whose height is 24.8 inches, and the radius of its base 30.6 inches? Ans. 25.73099 feet.

4. Required the solidity and convex surface of each of the frigid zones, whose height is 330.0075 miles, and the diameter of the earth 7957 miles.

Ans. The solidity is 1323679753,4398 cubic miles, and the convex surface is 8250209.7425 square miles.

PROBLEM XIII.

To find the solidity of the frustum or zone of a sphere. RULE.

To the sum of the squares of the radii of the two ends, add of the square of their distance, or the height of the zone; and this sum being multiplied by the said height, and the product again by 1.5708, will give the solidity.

Note. The convex surface of a zone may be found by multiplying the cir cumference of the whole sphere by the height of the zone.

EXAMPLES.

1. Required the solidity of the zone ABCD; the greater diameter AB being 28, the less diameter CD 20, and the height mn 15 inches.

+75=371; and 371 x 15 x 1.5708 = 8741.502 inches, the solidity required.

5565 x 1.5708

2. What is the solidity of a zone whose greater dia meter is 9 feet 3 inches, less diameter 6 feet 9 inches, and height 5 feet 6 inches? Ans. 370.32425 feet.

3. Required the solidity and convex surface of the torrid zone, the top and bottom diameters of which are each 7297.735 miles, its height 3173.14565 miles, and the circumference of the earth-25000 miles?

Ans. The solidity is 149455316338.46698 cubic miles, and the convex surface is 79328641.25 square miles.

PROBLEM XIV.

To find the solidity of a circular spindle.

RULE.

Find the area of the revolving segment ACBEA, which multiply by half the central distance OE. Subtract the product from of the cube of AE, half the length of the spindle; and multiply the remainder by 12.5664, and the product will be the solidity.

Note 1. The surface of a circular spindle may be found by the following rule. Multiply the length AB of the spindle by the radius OC of the revolv ing arc ACB. Multiply also the length of the said arc by the central distance OE, or distance between the centre of the spindie and the centre of the re