ULLAGE OF CASKS.

Art. 2. When a cask is partly filled, the whole capacity is divided, by the surface of the liquor, into two portions; the least of which, whether full or empty, is called the ullage. In finding the ullage, the cask is supposed to be in one of two positions; either standing, with its axis perpendicular to the horizon; or lying, with its axis parallel to the horizon. The rules for ullage which are exact, particularly those for lying casks, are too complicated for common use. The following are sufficiently near approximations. See Hutton's Mensuration.

PROBLEM VII.

To calculate the ullage of a standing cask.

RULE. Add together the squares of the diameter at the sur face of the liquor, of the diameter of the nearest end, and of double the diameter in the middle between the other two; multiply the sum by of the distance between the surface and the nearest end, and the product by .0028 for ale gallons, or .0034 for wine gallons.

If D=the diameter of the surface of the liquor,

d=the diameter of the nearest end,

m the middle diameter, and

l=the distance between the surface and the nearest end;

The ullage in inches (D2+d+2m2)×÷l×.7854.

Ex. If the diameter at the surface of the liquor, in a standing cask, be 32 inches, the diameter of the nearest end 24, the middle diameter 29, and the distance between the surface of the liquor and the nearest end 12; what is the ullage?

Ans. 27 ale gallons, or 333 wine gallons.

PROBLEM VIII.

To calculate the ullage of a lying cask.

RULE. Divide the distance from the bung to the surface of the liquor, by the whole bung diameter, find the area of a cir cular segment, whose versed sine is the quotient in a circle, whose diameter is 1, and multiply it by the whole capacity of the cask, and the product by 1 for the part which is empty,

If the cask be not half full, divide the depth of the liquor by the whole bung diameter, and find the area of the segment, multiply, &c., for the contents of the part which is full.

Ex. If the whole capacity of a lying cask be 41 ale gallons, or 493 wine gallons, the bung diameter 24 inches, and the distance from the bung to the surface of the liquor 6 inches, what is the ullage? Ans. 7 ale gallons, or 9 wine gallons.

OF THE

SPECIFIC GRAVITY OF SOLIDS AND FLUIDS.

THE specific gravities of bodies are their relative weights, contained under the same given magnitude as a cubic foot, or a cubic inch, &c.

A table of the Specific Gravities of Bodies, and the weight of a cubic foot of each, in ounces, avoirdupois.

Note. The several sorts of wood are supposed to be dry. Also as a cubic foot of water weighs just 1000 ounces, avoirdupois, the numbers in this table express not only the specific gravities of the several bodies, but also the weight of a cubic foot of each, in avoirdupois ounces; and hence, by proportion, the weight of any other quantity, or the quantity of any other weight, may be known as in the following problems.

PROBLEM I.

To find the magnitude of any body from its weight.

RULE. As the tabular specific gravity of the body is to its weight in avoirdupois ounces, so is one cubic foot, or 1728 cubic inches, to its content in feet, or inches respectively.

Ex. 1. Required the solid content of an irregular block of common stone, which weighs 1 cwt. or 1792 ounces. Here, as 2520 oz: 1792 ox:: 1728 cubic inches: to its solid content.

Or,

As 5 oz.: 256 oz :: 24 cubic inches: 1228 cubic inches, the solid content required.

Ex. 2. How many cubic feet are there in a ton weight of dry oak? Ans. 38 cubic feet. Ex. 3. What is the solid content, and diameter of a cast iron ball, that weighs 42 pounds, its specific gravity being 7208?

=

Solidity 161.095 cubic inches. Ans. Diameter = 6.75 inches.

PROBLEM II.

To find the weight of a body from its magnitude.

RULE. As one cubic foot, or 1728 cubic inches, is to th solid content of the body, so is its tabular specific gravity to the weight of the body.

Ex. 1. Required the weight of a block of marble, whose specific gravity being 2700, the length =63 feet, the breadth and thickness each 12 feet; this block being the dimensions of one of the stones in the walls of Balbeck.

=

Here, as 1 cubic foot: 63 x 12 x 12 (= 9072 cubic feet) :: 2700 oz. 683, tons weight, almost equal to the burthen of an East India ship.

Ex. 2. What is the weight of a block of dry oak, which measures 10 feet long, 3 feet broad, and 24 feet deep?

Ans. 4335 pounds. Ex. 3. What is the weight of a leaden ball, 4 inches in diameter, its specific gravity being 11351?

Ans. 16 lbs. 7 oz. Ex. 4. Required the weight of a cast iron shell, 3 inches thick, its external diameter being 16 inches, and its specific

gravity 7208.

2

Solidity 1621.0656 cubic inches. Ans. Weight = 2 cwt. 3 qr. 9 lb. .5 oz. {Solidity

PROBLEM III.

To find the specific gravity of a body heavier than water.

RULE.-Weigh the body both in water and out of water, by a hydrostatic balance, and take the difference of these results, which will be the weight lost in water.

Then say, as the weight lost in water, is to the weight of the body in air, so is the specific gravity of water, to the specific gravity of the body.

Ex. 1. A piece of stone weighed ten pounds in air; but, in water, only 6 pounds; required the specific gravity. Here, as 10-62 (31): 10: 1000 to the specific gravity of the body.

Or,

As 13 lbs.: 40 lbs. : : 1000 oz. : 3077 oz. = Ans. Ex. 2. A piece of copper weighs 36 oz. in air, and only 31.904 oz. in water; required the specific gravity of copper. Ans. 8788 ounces. Eq. 2. Required the specific gravity of a piece of granite stone which weighs 7 lbs. in air, and 5 lbs. in water. Ans. 3500 ounces.

PROBLEM IV.

To find the specific gravity of a body lighter than water.

RULE. - Fasten to the lighter body, by a slender thread, another body heavier than water, so that the mass compounded of the two may sink together. Weigh the heavier body, and the compound mass, separately, both in water and out of it, then find how much each loses in water, by subtracting its weight in water from its weight in air.

Then say, as the difference of these remainders is to the weight of the lighter body in air, so is the specific gravity of water to the specific gravity of the lighter body.

Ex. 1. Suppose a piece of elm weighs 12 lbs. in air, and that a piece of metal, which weighs 18 lbs. in air, and 16 lbs. in water, is affixed to it, and that the compound weight is 6 lbs. in water; required the specific gravity of the elm.

Here 18-16=2 pounds, the metal lost in water ;

and (18+15) 633-6=27 pounds, the compound lost in water.

Then 27-2=25 pounds, the elm lost in water.

2 (=25 lbs.): 15 lbs. : : 1000 oz. to the specific

gravity of the elm.

Or,

As 1 lb. : 3 lbs. :: 200 oz. : 600 oz. = Ans.

Ex. 1. A piece of ash weighs 20 lbs. in air, to which is affixed a piece of metal, which weighs 15 lbs. in air, and in water, 13 lbs.; and the compound, in water, weighs only 8 lbs; required the specific gravity of the ash.

Ans. 800 ounces.

Ex. 2. Suppose a piece of fir weighs 11 lbs. in air, and a piece of steel being affixed which weighed 16 lbs. in air, and in water 14 lbs. ; and the compound in water weighs only 5 lbs.; what is the specific gravity of the fir?

Ans. 550 ounces. Ex. 4. A piece of cork weighing 20 lbs. in air, had a piece of granite fixed to it, that weighed 120 lbs. in air, and 80 lbs. in water; the compound mass weighed 16 lbs. in water; what was the specific gravity of the cork?

Ans. 240 ounces.

QUESTIONS FOR EXERCISE.

1. Having a rectangular marble slab, 58 inches by 27, I would have a square foot cut off parallel to the shorter edge; I would then have the like quantity divided from the remainder, parallel to the longer side; and this alternately repeated, till there shall not be the quantity of a foot left; what will be the dimensions of the remaining piece?

Ans. 20.7 inches by 6.086. 2. Given two sides of an obtuse angled triangle, which are 20 and 40 poles; required the third side, that the triangle may contain just an acre of land? Ans. 58.876 or 23.099.

3. The ellipse in Grosvenor-square measures 840 links across the longest way, and 612 the shortest, within the rails; now the walls being 14 inches thick, what ground do they inclose, and what do they stand upon?

Ans.

{

stand on 1760 sq. feet. 4. What is the length of a chord, which cuts off one-third of the area, from a circle whose diameter is 289 ?

Ans. 278.6716.