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EXERCISES.

1. What is the content of a cask, whose bung diameter is 32 inches, end diameter 18, and length 38 inches?

Ans. 69.4.

2. Find the content of a cask, whose diameters are 40 and 20, and length 50 inches.

MEAN DIAMETERS OF CASKS.

Ans. 132.2.

167. The mean diameter of a cask is the diameter of a cylinder of the same length, whose capacity is equal to that of the cask.

The mean diameter may be found by means of the following table, the construction of which is this:-If the bung diameter be denoted by 1, the contents of the four varieties of casks will be expressed by

} (2+H2), (8+4 H+3 H2), 3 (1+H2), 3(1+H+H2), multiplied by 7854 L; but if D = the mean diameters, the contents are also expressed by 7854 D'L; hence, as each of the above expressions x 7854 L is equal to the last, therefore these expressions themselves are = D2 for the four varieties, or D for these varieties is equal to the square root of each of them. For example, when H = 5, B being = 1, then, for the first variety, D=√(2+H2) = {√3 =866; which is the number under the first variety in the following table opposite to 5 or H', which is just the value of H when B 1. In the same manner, all the other numbers in the table are found, for H, or rather H' = ·51, ·52, ·53,... up to 1. The numbers marked H' are just the ratio of the bung to the head diameter, and the numbers under the different varieties are the mean diameters when the bung diameter is = 1, and the ratio H' is its head diameter. For when B is assumed = 1, that is, as the unit of measure, the following proportion will exist, namely,

B:1-H:H' or H'H÷ B.

TABLE OF MEAN DIAMETERS, WHEN THE BUNG DIAMETER

IS= 1.

(The numbers are all decimal fractions, except those in the last line,
in which 1 is an integer.)

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8754

73 9188 9135 8685 99 9966 9966 9950 9950 74 9215 9166 8796 8732 100 10000 10000 10000 10000 75 9442 9196 8838 8780

168. PROBLEM XIII.-To find the capacity of a cask of any of the four varieties, by means of their mean diameters, found by the table.

'Divide the head by the bung diameter, and find the quotient in the column marked H' in the table, and opposite to it and under the proper variety is the mean diameter of a similar cask, whose bung diameter is 1.

Multiply this tabular mean diameter by the given bung diameter, and the product is the required mean diameter, the square of which, multiplied by the length, and divided by 353-036, is the content in imperial gallons."

=

If H' = ratio of H to D, and D' tabular mean diameter, then D' is found opposite to H', and then D = BD', C = D2L ÷ 353-036, or C =·0028325 D2L.

and

For, if D

the mean diameter, the content is

C=

.7854 277-274

· D2L = D2L÷ 353·036.

Instead of this divisor, the corresponding multiplier may be used, and then C0028325 D2L.

By the Sliding Rule.

'Set the gauge point 18-79 on D to the length L on C, then opposite to the mean diameter D on D is the content on C.

EXAMPLE. Find the content of a cask of the first variety, whose diameters are 30 and 24, and length 36.

H' =

then

H 24

B=308, and opposite to 8 is 938 = D';

D= BD' = 30 × 938 = 28·14,

and C0028325 D2L0028325 × 28·142 × 3680·7. Or, set 18-79 on D to 36 on C, then at 28.14 on D is 80.7 on C.

EXERCISE.-What are the contents of each of four casks of the four varieties, their diameters being 32 and 24, and length 40? Ans. The contents will be nearly the same as those in the former examples for the four varieties, and also the same as those in the first exercise of the next problem.

CONTENTS OF CASKS WHOSE BUNG DIAMETERS AND LENGTHS ARE UNITY.

169. The contents of casks may be more readily computed by means of a table of the contents of casks, whose bung diameters and lengths are = 1.

Let D' have the same meaning as in the preceding problem, that is, let it denote the numbers in the preceding table under the different varieties, which are just the mean diameters of casks, whose bung diameters are 1 and head diameters H'; also let C' be the content of a cask, whose

mean diameter is D' and length 1, and which the standard cask, then

*7854 277-274

may be called

C' = D'2L' X Hence, if the numbers in the preceding table are squared, and the square divided by 353-036, or multiplied by 0028326, the results will be the contents C' required. The following table can thus be constructed from the preceding one. Thus, for example, when H÷ B, or H' is 75, then, by the preceding table, D' for the second variety is 9196, and

= D'2÷353-036, for L' = 1.

C'D'2353·036 : =·9196 ÷ 353·036·0023959, which is just the capacity opposite to 75 in the following table.

TABLE OF CONTENTS IN IMPERIAL GALLONS OF STANDARD

CASKS C'.

H'

First Variety. Second Variety. Third Variety. Fourth Variety.

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H'

First Variety. Second Variety. Third Variety. Fourth Variety.

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170. PROBLEM XIV.-To find the content of a cask by means of the table of contents of standard casks.

'Divide the head by the bung diameter, and find the quotient in the column H', and opposite to it and under the proper variety is the content C' of the standard cask; multiply this tabular content by the square of the bung diameter of the given cask, and this product by the length, and the result will be the required content in imperial gallons.'

H' HB, opposite to H' is C', then C = C'B2L.
For by 168, CD2L 353.036,

and D2 = B2D'2, also (169) C′ = D'2 ÷ 353·036;
hence,

C=C'B2L.

EXAMPLE. Find the content of a cask of the first variety, whose diameters are 30 and 24, and length 36.

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