Εικόνες σελίδας
PDF
Ηλεκτρ. έκδοση

in. wide at the edge; and 6 in. thick at the large

[blocks in formation]

864 product.

Pro. 864X 48 41472-66912 in. or 4 ft. Ans.

2. What is the solidity of a wedge that is 20 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

Width of large end 9×2

0

18

18 reserved sum.

21 X9189X183402÷6567 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 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. Cube the diameter, or axis, and multiply its cube by 5236 the last product is the solidity.

Examples.

1. What is the solidity of a globe, or sphere whose diameter is 1137

#113×11 3×113 × ·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 ? Ans. 381.7044 sol. in. 4. What is the solidity of a sphere whose diameter is 2 ft. 8 in. 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 G F.

[blocks in formation]

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

Examples.

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

Radius 7X7X3
Depth 5X5

147 reserved product.

25 add.

172× depth 5860×·5236450.296 Ans.

S

2. Required the solidity of a segment of a globe whose semidiameter, is 9 in. depth 9 in.

Ans. 1526.8176 in.

CASE XII.

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 figure in case eleventh.

RULE.-Square the semidiameter 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.

Examples.

1. What is the solidity of the middle zone C DEF, (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.

Diam. 14 semi.=7x7=49
Diam. 14 semi.=7x7=49

Height o M 3× 3 =9÷ 3 add.

ht.

Ans.

101X3=303X1.5708-475-9524

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. 1340-41†.

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 a solid generated by the revolution of an ellipse, or oval, about the transverse or conjugate diameter. [See the figure A B C D.]

[graphic][subsumed][subsumed][subsumed][subsumed]

To find the solidity of the spheroid A B C D.

RULE Multiply the revolving diameter B D, into it self, and the product by the fixed diameter A C, and the product again by 5236, the last product is the solidity.

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=400x30x5236-6283.2 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. To the square of the end diam. 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.

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.?

[ocr errors]

End diam. 24×24= 576

Mid. diam. 32×32×2=2048 add.

2624×40X 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.-An elliptic spindle is formed by any of the three conic sections revolving about a double ordinate; the following figure represents an elliptic spindle.

[blocks in formation]

RULE. Find a diameter half way between the middle and end, multiply it by two, square the product and add the square to the square of a diameter taken in the middle, multiply the sum by the length, and the product again by 1309, the last product is the solidity, very nearly.

Examples.

1. What is the solidity of the elliptic spindle VBCD; length 15; diameter C D 5; and diameter taken half way between the middle and end equal to 4? Diam. O P. 4x2=8×8=64 sq. of twice O P. Diam. C D 5×5=25 sq. of C D add.

89 sum.

Sum. 89x15x-1309—1747515 Ans.

2. What is the solidity of an elliptic spindle thar is 22 long; greatest diameter 15; and the diameter har between the middle and end, equal to 10? P

Ans. 189025

« ΠροηγούμενηΣυνέχεια »