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of the revolving arch; subtract this last product from the fore mer, and multiply the remainder by 6.2832 for the superficies.

EXAMPLE. Required the area of a circular spindle, whose length is 40 and thickness 30 inches.

a

AD2+BD2 = AB the chord of the arch ABC; that is,

400+225 = 25 AD2

=DH and DH+BD=FB rad. also FB-BD=DF cent. dift. BD

2

400

= 26.6 and 26.6+16=20.83 rad. also 20.83-15=5.83=DF 15

2

Now, to find the length of the arch:

As AF = 20.83 = 1.31869
Is to rad. 90

10.00000
So is AD=20

1.30103

To Sine arch 73° 41'=9.98234.

2

147° 22' arch.

Then fay, As 360°: 147° 2 :: 3.1416X41.6: 53.58 leng. of arche

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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. 2747.3166336. 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 folidity 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 folidity. See the laft figure.

EXAMPLE I.

Required the area of a circular spindle, wltose length is 60 and diameter 43 Dd

AD

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BD +AD: = AB = 506.25 +900 = 37.5

37.5 chord half arch.

8

300.0
60 chord whole arch.

3)240

80 the length of the arch.

31.25 rad.

80 length of the arch. 2)2500.00

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

987.5 the area of the rev. segm. ACB. 4.375 the half the central dift. DF,

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49375 69125 29625 29500

4320.3125

30 half the spindle.
30

900
30

3)27000

gooc one-third cube { fpindle. 4320.3125

4679.6875

4

18718.7500

3.1416

1123125000
187187500
748750000
187187500
561572500
58806.82500000 solidity.

Ex. 2. Required the folidity of a circular spindle, whose length is 30, and thickness 22 inches. Anf: 7350.853125.

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

Anf. 29919} cubic inches.

PROBLEM XVII.

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

RULE.

From the fourth part of the square of the length of the whole spindlę, subtract the square of half the length of the middle frustum, 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 EF: +FB2=EB?=500 chord. EB2

500

--= 50 diameter of the generating circle. FB

10

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

7
AL-=AD2-LD-=625-49=576
EF= 400 =

133.3
3 3

442.6

20

8853-3 first product.

BE

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