Note. The use of the Multipliers, Divisors, and Gauge-points, is shown in the

PROBLEM I.

The side of a vessel in the form of a cube, being given in inches; to find its content in imperial gallons and bushels.

RULE.

Multiply the side by itself, and that product again by the side; and the last product will be the content of the vessel in cubic inches; which being multiplied by .003607, and .000451; or divided by 277.3, and 2218.2, the respective products or quotients will be the content of the vessel in imperial gallons and bushels.

Note 1. The definitions of the cube, the parallelopipedon, the cylinder, &c. &c. may be seen in Part IV.; and if the content of any vessel be found in cubic inches, by the rules given in that Part, and then divided by 282, 231, and 2150.4, the respective quotients will be the content of the vessel in old ale and wine gallons, and malt bushels; but the Rules and Examples given in the following Problems will be found more particularly adap ed to Gauging, than those in Part IV.

2. If the content of any vessel, in cubic inches, be divided by 277.3, and 2218.2; the respective quotients will be the content in imperial gallons and bushels.

3. Gaugers always take their dimensions in inches or in inches and tenths; and when we say the side or diameter of a vessel measures so many inches, we mean the internal, not the external dimensions.

EXAMPLES.

1. The side of a cubical vessel measures 46 inches; what is its content in imperial gallons and bushels?

Here 46 × 46 × 46=2116 × 46=97336, the content in cubic inches; then 97336 ×.003607=351.0909 imperial gallons; and 97336 ×.000451=43.8985 imperial bushels. Or, 97336÷277.3=351.0133 imperial gallons; and 97336÷2218.2=43.8806 imperial bushels.

Note. The divisors are used in all the following solutions; the work, however, may be proved by the multipliers given in the preceding Table.

By the Sliding Rule.

In this operation, the square gauge-points 16.65, and 47.10, upon the line D, must be used.

Note. New Sliding Rules have not always the square guage-points marked upon them; but small brass pins may be easily inserted at those points.

2. The side of a cubical cistern is 134 inches; what is its content in imperial gallons and bushels?

Ans. 8676.8986 imperial gallons, and 1084.7101 imperial bushels.

3. The side of a cubical wine-vat measures 135.6 inches; how much wine will it contain at once?

Ans. 8991.4389 gallons

PROBLEM II.

8991 gal. 1 qt. 1 pt.

The length, breadth, and depth of a vessel, in the form of a parallelopipedon, being given in inches; to find its content in imperial gallons and bushels.

RULE.

Multiply the length by the breadth, and the product thence arising by the depth; and the last product will be the content in cubic inches; which being divided by 277.3 and 2218.2, will give the content in imperial gallons and bushels.

Note. As the sides of vessels, in the form of a parallelopipedon, are seldom perfectly regular and parallel, it is best to take several lengths, in different places; and divide their sum by their number for a mean length. A mean breadth and depth may be found in the same manner

EXAMPLES.

1. A cistern in the form of a parallelopipedon, measures 96 inches in length, 65 in breadth, and 48 in depth; what is its content in imperial gallons and bushels ?

Here 96 × 65 × 48=299520, the content in cubic inches, then 299520-277.3=1080.1298 gallons; and 299520 ↓ 2218.2=135.0284 bushels.

By the Sliding Rule.

As 96 on C: 96 on D::65 on C: 79 on D, which is a mean proportional between the length and breadth. (See Problem 5, page 152.) Then,

2. The length of a vessel in the form of a parallelopipedon, is 136, its breadth 94, and its depth 62 inches ; what is its content in imperial gallons and bushels?

Ans. 2858.3050 gallons, and 357.3203 bushels. 3. A water-trough measures 85.3 inches in length, 54.7 in breadth, and 32.9 in depth; how many imperial gallons will it contain?

4. A maltster has a cistern whose length is 132, breadth 118, and depth 46 inches; how much barley can he steep at a time, admitting the water to occupy of the cistern? Ans. 193.8047 bushels = 24 qr. 1 bush. 31 pk.

PROBLEM III.

The diameter and depth of a vessel, in the form of a cylinder being given; to find its content in imperial gallons and bushels.

RULE.

Multiply the square of the diameter by the depth; and divide the product by 353, for imperial gallons; and by 2824.3, for imperial bushels.

Note 1. As cylindrical vessels are seldom perfectly round, it is best to measure cross diameters, in different parts; and divide their sum by their number, for a mean diameter.

2. If a cylindrical vessel be placed in an inclining position, so that the liquor forms an elliptical surface, the content may be found by the following Rule: Multiply the square of the diameter of the vessel by half the sum of the greatest and least depths of the liquor; and divide the product by 353, for imperial gallons. (See Notes, Prob. 5.)

EXAMPLES.

1. The diameter of a cylindrical vessel is 34, and its depth 45 inches ; what is its content in gallons and

bushels?

Here 34 x 34 × 45 = 52020; then 52020÷353= 147.3654 gallons; and 52020÷2824.3=18.4187 bushels.

By the Sliding Rule.

Here the circular gauge-points 18.79, and 53.14, upon the line D, must be used.

2. The diameter of a cylindrical vessel is 46.7, and its depth 68.4 inches; what is its content in gallons and bushels? Ans. 422.5860 gallons, and 52.8176 bushels.

3. A cylindrical mash-tun measures 94 inches in diameter, and 82 in depth; how many bushels of malt will it contain at once? Ans. 256.5421 bushels. 4. At Heidelberg, in Germany, is a cylindrical wine cask,

whose depth is 27, and diameter 21 feet; how many gallons will it contain, imperial measure?

Ans. 58286.9575 gallons.

Note. The convivial monument of ancient hospitality, mentioned in the 4th Example, was formerly kept full of the best Rhenish wine, and the electors have given many entertainments on its platform; but it now only serves as a melancholy instance of the extinction of that hospitality; for it is suffered to moulder in a damp vault, quite empty.

Although this vessel is of an extraordinary magnitude, yet it is much inferior in capacity, to many of the London porter-vats

PROBLEM IV.

Given the dimensions of a vessel in the form of a prismoid, or the frustum of a square pyramid, or a cylindroid; to find its content in imperial gallons and bushels.

RULE.

To the sum of the areas of the two ends, add four times the area of the middle section parallel to them; multiply this sum by the perpendicular depth, and of the product will be the content in cubic inches; which divide by 277.3 for imperial gallons, and 2218.2 for imperial bushels. (See the Scholium, Prob. 10. Sect. I. Part IV.)

Note 1. A cylindroid is a figure resembling the frustum of a cone; but having elliptical instead of circular ends. Sometimes one end is circular, and the other elliptical.

2. When the vessel is a prismoid, the length of the middle section is equal to half the sum of the lengths of the two ends; and its breadth is equal to half the sum of their breadths.

3. If the ends be elliptical, the transverse diameter of the middle section will be equal to half the sum of the transverse diameters of the two ends; and the conjugate diameter equal to half the sum of the conjugate diameters of the two ends.

4. If one end be an ellipse and the other a circle, add the transverse diameter of the elliptical end to the diameter of the circular end; and take half the sum for the transverse diameter of the middle section. The conjugate diameter of the middle section may be found in a similar manner; it is better, however, in all cases of practice, to take the real dimensions of the sections.

5. When the ends are rectangles, their areas may be found by Problem 2; when they are circles, we may obtain their areas by Problem 15; and when they are ellipses, we can find their areas by Problem 21, Part II.

EXAMPLES.

1. The length and breadth of the bottom of a vessel in the form of a prismoid measures 72 and 64, the length and breadth of the top 96 and 82, and the perpendicular depth 65 inches; what is its content in gallons and bushels?

Here (72 × 64) + (96 × 82)=4608+7872=12480, the area of the two ends.

Also, (7296)+2=168÷2=84, the length of the