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Ex. 2. Required the number of square inches which will cover a circular spindle, whose length is 80 and thickness 16 inches ?

Anf. 2747.3166 336. Ex. 3. Required the area of a circular spindle, whose length is 12, and thicknefs y inches.

Anf. 294.3621 sq. inches.

PROBLEM XVII.

To find the solidity of a circular spindle.

Multiply the area of the revolving segment by the distance between the centres of the arch and spindle, subtract the prcduct from the cube of half the length of the spindle, then multiply the remainder by 4, and this product again by 3.1416 for the solidity. See the last figure.

EXAMPLE I.

Required the area of a circular spindle, whose length is 60 and diameter 4). Dd

AD

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1250 the area of the sector AFCB. 8.75 X 30= 262.5 the area of the triangle ACF.

937.5 the area of the rev. segm. ACB.
4.375 the half the central ditt. DF.

49375 60125 29625 29500

4320.3125

30 half the spindle.
30

900
30

3)27000

9000 one-third cube į spindle.
4320.3125
4679.6875

4

18718.7500

3.1416

1123125000
187187500
748750000
187187500
561572500
58806.82500000 folidity.

Ex. 2. Required the solidity of a circular spindle, whose length is 30, and thickness 224 inches. Anf. 7350.853125 Ex.

3. Required the solidity of a circular spindle, whose middle diameter is 36, and length 40 inches.

Anf. 29919 citbic inches.

PROBLEM XVIII.

To find the folidity of the middle zone of a circular spindle.

RULE.

From the fourth part of the {quare of the length of the whole spindle, subtract the square of half the length of the middle fruitum, and multiply the remainder by the length

of

of the frustum : Multiply the central distance by the revolving area which generates the frustum ; then subtract this latter product from the former, and multiply the remainder by 3.1416, and twice the product will be the solidity.

EXAMPLE I.

Required the folidity of the frustum of a circular spindle, whose length is 40, greatest diameter 36, and least 16 inches.

Draw EG parallel to mn, then EF shall be equal , mn, =20

and EF2 +FB-=EB=500 chord. EB2

500

-= 50 diameter of the generating circle. FB

10

Hence rad. BD = 25
and 25--18 = 7 the central dift.

AL-=AD2-LD-=625—495576
EF=
= 400 =

133.3

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3 3

442.6

20

8853.3 first product.

BE

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Ex. 2. Required the folidity of a circular spindle, whose length is 40, its greatest diameter 32, and least 24 inches.

Anf. 27287 cubic inches.

PROBLEM

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