EXAMPLES.

1. What is the solidity of a wedge that is 12 ft. 1ong, 12 wide, and 6 thick, at the large end?

12X12=144×6÷3-432 solidity, Ans. 2. What is the solidity of a wedge that is 10 ft. long, 6 inches wide, and 6 thick? Ans. 14 ft.

CASE IX.

To find the solidity of a wedge when the edge is narrower than the large end.

RULE. (1) To the width of the edge add twice the width of the large end, and reserve the sum: multiply the length of the wedge by the thickness of the large end; multiply this last product and the reserved sum together, and divide by 6; the quotient is the solidity.

Question.-How do you find the solidity of a wedge, when the edge is narrower than the large end?

EXAMPLES.

1. What is the solidity of a wedge that is 12 ft. long, and 18 in. wide at the largest end, and 12 in. wide at the edge; and 6 in. thick at the large end?

Width of the edge

12

864 product.

Pro. 864X48-41472+6=6912 in. or 4 ft. Ans.

2. What is the solidity of a wedge that is 20 ft. long, 15 wide at the large end, and 10 at the edge; and 12 thick? Ans. 1600.

NOTE. This rule will give the solidity of a cone when it is square, or in form of a parallelogram.

EXAMPLE.

What is the solidity of the cone, figure third, case fifth; width of the large end 9, thickness 9, and length 21?

Width of the edge

0

Width of the large end 9X2=18

18 reserved sum.

21X9=189×18-3402-6-567 Ans.

(See Ans. to Quest. 3 Case fifth.)

CASE X.

To find the solidity of a globe, or sphere.

DEFINITION. A globe or sphere is (2) a round body bounded by a surface every point of which is equally distant from a point within called the centre; a line passing from one side to the other through the centre is called the diameter, or axis.

RULE. (2) Cube the diameter, or axis, and multiply its cube by 5236 the last product is the solidity.

Questions.-1. What is a globe or sphere? 2. What is the rule for finding the solidity of a Globe or Sphere.

EXAMPLES.

1. What is the solidity of a globe, or sphere whose diameter is 113?

113×113×113× 5236-7355500-8692 solidity.

2. What is the solidity of the globe which we inhabit, in solid miles; allowing its circumference to be 25000 miles? As 22 is to 7 so is 25000 to 7954 diam. nearly. (Fractions omitted.) 263485304337 sol. miles. Ans.

3. What is the solidity of a cannon ball that is 9 inches on its diameter?

4. What is the solidity of a 2 ft. 8 in.?

Ans. 381 7044 sol. in. sphere whose diameter is Ans. 9.92† ft. solidity.

CASE XI.

To find the solidity of a segment, or part of a globe, or

sphere; or part of a globe cut off parallel

to the diameter, as part D C F.

See the figure.

RULE. (1) Square the radius of its base, (as D o, or Fo) multiply its square by 3, and reserve the product; square the depth o C, add the square and reserved product together; and multiply the sum by the depth o C; and the last product by 5236; the product is the solidity.

Question.-1. What rule can be given for finding the solidity of a segment or part of a globe or sphere.

EXAMPLES.

1. What is the solidity of the segment D C F, radius Do, or F o, 7; and depth o C, 5?

Radius 7×7×3= 147 reserved product.

Depth 5X5 25 add.

172xdepth 5=860X·5236=450-296. Ans.

3. Required the solidity of a segment of a globe, whose semi-diameter is 9 in. depth 9 in.

CASE XII.

Ans. 1526-8176 in.

To find the solidity of the middle zone of a sphere, or globe; or the part of a sphere after two segments have been cut off, parallel to the diameter or

axis, as C D E F in figure, case eleventh.

RULE.-() Square the semi-diameter of both ends and add the squares together, and reserve the sum: square the height, or distance of the two ends, as o M, and add of its square to the reserved sum; multiply the sum by the height, or distance o M, and this product again by 1.5708; the last product is the solidity.

Question.-1. How do you find the solidity of the middle zone of sphere or globe?

EXAMPLES.

1. What is the solidity of the middle zone C D E F, (case eleventh;) diameter C E, or D F, 14, and height o M, 3?

NOTE. In this example the diameters are alike, which is not always

the case.

101 X3 303X1.5708-475-9524. Ans.

2. What is the solidity of the middle zone of a sphere whose greatest diameter is 12, and least 8, and height or length 10? Ans. 1840-41†

CASE XIII.

To find the solidity of a spheroid, or ellipsoid; and also to find the solidity of the middle frustum of a spheroid. DEFINITION.-A spheroid is (1) a solid generated by the revolution of an ellipse, or oval, about the transverse of conjugate diameter. [See the figure A B C D.]

itself, and the product by the fixed diameter A C, and the product again by ·5236, the last product is the solidity. Questions.-1. What is a spheroid?-2. How do you find the solidity of a spheroid?

EXAMPLES.

1. What is the solidity of a spheroid whose revolving diameter B D is 20, and fixed diameter A C is 30? Revolv. diameter 20×20=400×30×·5236=6283-2 Ans. 2. What is the solidity of a spheroid whose revolving diameter is 30, and fixed diameter 50? Ans. 23562.

To find the solidity of the middle frustum of a spheroid. RULE.-(1) To the square of the end diameter, add twice the square of the middle diameter, multiply this sum by the length, and the product again by 2618, the last product is the solidity.

Question.-1. How do you find the solidity of the middle frustum of a spheroid?

EXAMPLES.

1. What is the solidity of the frustum E F G H, length 40, end diameters E G, or F H 24, and middle diameter BD 32?

2624×40× 2618=27478.5280 Ans.

2. What is the solidity of the middle frustum of a spheroid, whose length is 30; middle diameter 25; and end diameter 20 inches? Ans. 12959.1.

CASE. XIV.

To find the solidity of an elliptic spindle. DEFINITION. (1) An elliptic spindle is formed by any of the three conic sections revolving about a double ordinate; the following figure represents an elliptic spindle.