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And set 40 on C, to 59.41 on D; then against

8 on D, stands 0.7
56 on D, stands 35.6

35.6
35.6

sum 107.5 wine gall. Ex. 2. What is the content of a conical cask, the length being 20, and the bung and head diameters 16 and 12 inches?

10.985 ale gallons. Ans.

13.418 wine gallons.

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PROBLEM XVI.

To find the content of a cask by four dimensions. Add together the squares of the bung and head diameters, and the square of double the diameter taken in the middle between the bung and head : then multiply the sum by the length of the cask, and the product again by .00043 for ale gallons, or by .0005; for wine gallons.

EXAMPLES.

1. Required the content of any cask whose length is 40, the bung diameter being 32, the head diameter 24, and the middle diameter between the bung and the head 281 inches.

57.5

24

32 57.5

32

24

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BY THE SLIDING RULE. Set 40 on C, to 46.4 on D; then against

24 on D, stands 10.5 32 on D, stands 19.0. 573 on D, stands 62.0

sum 91.5 ale gallons. Set 40 on C, to 24.0 on D; then against

24 on D, stands 13.0 32 on D, stands 23.2 571 on D, stands 75.0

sum 111.2 wine gallons. Ex. 2. What is the content of a cask, whose length is 20, the bung diameter being 16, the head diameter 12, and the diameter in the middle between them 143 ?

11.4479 ale gallons. Ans.

{ 13.9010 wine gallons.

PROBLEM XVII. To find the content of any cask from three dimensions only.

Add into one sum 39 times the square of the bung diameter, 25 times the square of the head diameter, and 26 times the product of the two diameters: then multiply the sum by the length, and the product again by 00034

.00034 for wine gallons, or by or .00003 1., for ale 9

11 gallons.

EXAMPLES.

1. Required the content of a cask, whose length is 40, and the bung and head diameters 32 and 24. 32 24

32
32
24

24

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2972160
.00034

2972160

:.00003 i
11888640
8916480

8916480

270196 9)1010.53440

91.86676 ale gall. 112.2816 wine gall. Ex. 2. What is the content of a cask, whose length is 20, and the bung and head diameters 16 and 12?

S 11.4833 ale gallons.
Ans.

14.0352 wine gallons.

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Note.—This is the most exact rule of any, for three dimensions only; and agrees nearly with the diagonal rod.

OF THE ULLAGE OF CASKS.

The ullage of a cask is what it contains when only partly filled. And it is considered in two positions, namely, as standing on its end with the axes perpendicular to the horizon, or as lying on its side with the axes parallel to the horizon.

PROBLEM XVIII.

To find the ullage by the Sliding Rule. By one of the preceding problems find the whole content of the cask. Then sct the length on N, to 100 on SS, for a segment standing, or set the bung diameter on N, to 100 on SL, for a segment lying; then against the wet inches on N, is a number on SS or SL, to be reserved.

Next, set 100 on B, to the reserved number on A ; then against the whole content on B, will be found the allage on A.

EXAMPLES.

1. Required the ullage answering to 10 wet inches of a standing cask, the whole content of which is 92 gallons, and length 40 inches. Having set 40 on N, to 100 on SS; then against 10 on

N, is 23 on SS, the reserved number. Then set 100 on B, to 23 on A; and

against 92 on B, is 21.2 on A, the ullage required. Ex. 2. What is the ullage of a standing cask, whose whole length is 20 inches, and content 111 gallons; the wet inches being 5?

Ans. 2.65 galls. Ex. 3. The content of a cask being 92 gallons, and the bung diameter 32, required the ullage of the segment lying when the wet inches are 8.

Ans. 16.4 galls.

PROBLEM XIX.

To ullage a standing cask by the pen. Add all together, the square of the diameter at the gur. face of the liquor, the square of the diameter of the nearest end, and the square of double the diameter taken in the middle between the other two; then multiply the sum by the length between the surface and nearest end, and the product again by .00043 for ale gallons, or by .00054 for wine gallons, in the less part of the cask, whether empty or filled.

EXAMPLE

The three diameters being 24, 27, and 29 inches, required the ullage for 10 wet inches.

24 29 54
24 29 54 2916

841
96 261 216

576 48 58 270

4333 576 841 2916

10

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To ullage a lying cask by the pen. Divide the wet inches by the bung diameter, find the quotient in the column of versed sines, in the table of circular segments at page 231 of the book, taking out its corresponding segment. Then multiply this segment by the whole content of the cask, and the product again by 1} for the ullage required, nearly.

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