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EXAMPLE.

Suppose the figure ABFE to represent a copper, and ExFgE its rising crown, AB=90, CD=82, Ae=5, it is plain that AB~2 * Ae=90—10=EF, Ee=30, and Gg=27, confequently gx=3

To cover the crown.

Gall. In the core CDFE, the diameter ?

B F. Gall. CD=82, EF-30, and gx=3, its con

54.782 = 1 2 3.778 tent in gallons, &c. is

The base diameter 80, and height of the crown 3, its content in ale 26.737 = 0 3 1.235 gallons, &c. will be The liquor that will cover the crown is

28.945 = 0 3 2.543

The crown being thus covered, it now remains to find the content of the copper from the crown upwards, the depth being 27 inches. In order to this, take the diameter in the middle of every 10 inches from the top, and infert each against the parts of the depth, as in the following table. Find the area of each in ale gallons, by Problem IV. and insert these areas, each against its corresponding diameter, as in column third; alla the contents of the several parts of the depth are placed in the fourth column. And these contents, being reduced to barrels, firkins, gallons, are inserted in the last column, as follows.

Parts

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The content being thus found, you may proceed to inch the copper by the fame directions which were given for inching tuns in last problem.

CASK-GAUGING.

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CASK-GAUGING is the most difficult part of the art: This difficulty arises from the variety of curves which vessels may be composed of.

It is also the most imperfect, and ever will be; because no calk can be made in such strick conformity to the solid it represents, as by the rules of art it is required to be.

Gaugers have reduced all kinds of calks to four forms, or varieties.

Variety I. The middle frustum of a spheroid --Fig. 2.
Variety II. The middle fruftum of a parabolic spindle-Fig. 3.
Variety III. The middle frustun:s of two parabolicc onoids-

Fig. 4.

Variety IV. The middle fruftums of two cones-Tig. 5.

PROBLEM

PROBLEM IX.

To find the content of a cask,

RULE I.

For Variety I. To the square of the head diameter add twice the square of the bung diameter, then multiply the sum by the length, and divide the product by 1077 for ale gallons. and by 882 for wine gallons.

RULE II.

For Variety II. To g times the square of the bung diameter add 6 times the square of the head diameter, then multiply the sum by the length; divide the product, as above, for ale and wine gallons.

RULE III.

For variety III. To the sum and half sum of the squares of the head and bung diameters, add of the difference of their squares, then multiply the sum by the length, and divide the product, as above, for alé and wine gallons.

RULE IV.

For variety IV. From the sum and half sum of the squares of the head and bung diameters, subtract half the square of their difference; then multiply the remainder by the length, and divide the product, as above, for ale and wine gallons.

A general Rule for reducing caks to a cylinder.

First, consider which of all the four Varieties the proposed caík resembles, then from the bung diameter fubtract the head diameter, and multiply the difference by .7 for the spheroid, by .65 for the spindle, by •6 for the conoids, and by :55 for the cones ; add the product to the head diameter, the sum is a mean diameter, or the diameter of a cylinder of equal content and length with the cask proposed.

EXAMPLE.

The length of a cask is 40 inches, bung diameter 32, and head diameter 24 inches; required its content in ale and wine gallons.

Case I. Suppose the cask of the first form; then,

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29.6 x 29.6 x.00278 X 40=97.6 ale gallons. 29.6 % 29.6x.0034 X 40=119.1 wine gallons.

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By the sliding rule.

Set the length of the calk in inches on C to the gauge-point on D, and against the mean diameter on D you have the cone tent on C.

Case II. Suppose the cask of the second form; then,

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84.65=5.2 and 24+5.2=29.2 the mean diameter.
29.2 K 29.24.00278 x 40=94.98 ale gallons.
29.2 X 29 25.0034 X 40-115.959 wine gallons.

CASE

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