Imágenes de páginas
PDF
EPUB

breadth of the zone, and this sum multiplied by the said breadth, and the product again by 1.5708, will give the solidity.

EXAMPLES.

1. What is the solid content of the zone ABCD, whose greater diameter AB is 20 inches, the less diameter CD 15 inches, and the distance nm of the two ends 10 inches?

[merged small][ocr errors]

102 Here 103 +7.53 +-3)*10* 1.5708=(100+56.25 + 33.33) x 10 x 1.5708 189.58 x 10x1.5708 = 1895.8 x 1.5708=2977.92264 solid inches, the answer.

2. What is the solid content of a zone, whose greater diameter is 24 inches, the less diameter 20 inches, and the distance of the ends 4 inches ? Ans. 1566.6112 inches.

3. Required the solidity of the middle zone of a sphere, whose top and bottom diameters are each 3 feet, and the breadth of the zone 4 feet?

Ans. 61.7848 feet.

H

RE十m and exterminating H and k by the means of the two equations +

a2 pa and u—h=a,we shall have (r? +92 + 3) x 2,

which is the h rule.

If it be the middle zone of the sphere, the solidity will be=d? + 18) *.7854h ; where d=diameter of each end, and h=its height.

PROBLEM XVI.

To find the solidity of a spheroid.

RULE.*

Multiply the square of the revolving axe by the fixed axe, and this product again by .5236, and it will give the solidity required.

Where note that .5236 is=1 of 3.1416.

EXAMPLES.

1. In the prolate spheroid ABCD, the transverse, or fixed axe AC is 90, and the conjugate, or revolving axe DB is 70: what is the solidity ?

* Demon. Let Ac=, DB=b, Ar=x, rn=y, and p= 8.14159, &c.

62 Then a':b? :: x*(a—2): *(ax—*)=y by the property of the ellipsis.

And therefore the fluxion of the solid (=py?:)=P** (axm:); and its fluent x pb*

x (fax? — $x3)=segment

pabi NAM. Which, when x=a becomes

ax)

* (ļa—£a*)=7 =content of the whole spheroid. Q. E. D,

If f be put=fixed axe, r=revolving axe, q=\f02 gue); f?, and p=3.1416, &c:

Then will prf vi+tq=surface of the oblate spheroid and prf1-q=that of the prolate spheroid.

[ocr errors]
[merged small][merged small][ocr errors][merged small][merged small][merged small]

B Here DBP X ACX.5236=702 X 90 X.5236=4900 x 90 x .5236=441000 X.5236=230907.6=solidity required.

2. What is the solidity of a prolate spheroid, whose fixed axe is 100, and its revolving axe 60 ? Ans. 188496.

3. What is the solidity of an oblate spheroid, whose fixed axe is 60, and its revolving axe is 100 ?

Ans. 314160.

PROBLEM XVII. To find the content of the middle frustrum of a spheroid, its length, the middle diameter, and that of either of the. ends, being given.

CASE I. When the ends are circular, or parallel to the revolving

axis.

RULE.*

To twice the square of the middle diameter add the square of the diameter of either of the ends, and this sum

[ocr errors]

* Demon. Let Ao=a, Do=b, EN=h, no=c, ro=x, re= y, and p=3.14159, &c.

62 Then a? : 53 :: 22-02 : -X(a2-x2)=62

a2

aa a by the property of the ellipsis.

bax (as) And also a? : 63 :: ac2:

=h?; or a = bc-(-h.)

multiplied by the length of the frustrum, and the product again by .2618, will give the solidity.

Where note that .2618=1 of 3.1416.

EXAMPLES.

1. In the middle frustrum of a spheroid EFGH, the middle diameter DB is 50 inches, and that of either of the ends EF or GH is 40 inches, and its length nm 18 inches : what is its solidity ?

D

H

[ocr errors][merged small][merged small]

Here (509x2 +40%) x 18x.2618= (2500 x 2 +1600) 18x.2618= (5000 + 1600) X 18x.2618 = 6600 x 18x 2618 = 118800 x .2613= 31101.84 cubic inches, the

answer.

2. What is the solidity of the middle frustrum of a prolate spheroid, the middle diameter being 60, that of either of the two ends 36, and the distance of the ends 80?

Ans. 177940.224.

Whence, by substituting this value of a’ in the former

64x4-6%h%29 equation, we shall have y'=62

b?c box? h?? 22

-= ca

=B- (xb?—h.) And consequently the fluxion of the solid (py :)=pbirpx:

рх ;

- Х

3c2 (69-79); which, when x=c, becomes pbc

pebe - pche

3 pcx (262 +h) pc 3

=12*86+40°. Q. E. D.

3." What is the solidity of the middle frustrum of an oblate spheroid, the middle diameter being 100, that of either of the ends 80, and the distance of the ends 36 ?

Ans. 248814.72

CASE II.

When the ends are elliptical or perpendicular to the re

volving axis.

RULE.* 1. Multiply twice the transverse diameter of the middle section by its conjugate diameter, and to this product add the product of the transverse and conjugate diameters of either of the ends.

ex (

* Demon. Put so=a, Eo=b, om=r, on=X, AN=Y, XC =%, and p=3.14159, &c.

62 Then a

: b2 :: a*: x(a)=by the property of the ellipsis. And, since Acd is an ellipsis similar to EMF, it will be

ry bir::y:

b

=%; as is shown by the writers on Conics. But the Aluxion of the solid aefd is pyză = pyž x ry pry': pri box (a?—«?) 6 b

prbex

And the a?

a?

[ocr errors]

fluent=prbx x

a

Which, by substituting for aits 62x2

25+y

ry value €7--23

becomes = prxx
'

36
=pxxrb+

36 • And this again, by putting z for its equal **, becomes px

px ne - x 2rb+yz frustrum EFDA. Or X (2EFX 20m+

12 AD X 2ne)=middle frustrum ABCD. Q. E. D.

« AnteriorContinuar »