CASE 10.-To find the solid content of a Sphere or Globe.

NOTE. For definition of a Globe, see Case 9 of Superficies.

RULE. Find the superficial content by Case 9 of Superficies; multiply this surface by one-sixth of the diameter, and it will give the solidity.

Or, multiply the cube of the diameter by ,5236-and the product will be the solidity.*

EXAMPLE.

What is the solidity of our earth, if its diameter be 7957 miles, nearly, and its circumference at the equator be just 25000 miles?

7957,75×25000×7957,75÷6-263857106187,5+solid

miles. Ans.

CASE 11.-To find the solid content of a frustum or segment of a Globe.

Definition. The frustum of a globe is any part cut off by a plane.

RULE. To three times the square of the semidiameter of the base, add the square of the height; multiply this sum by the height, and the product again by ,5236; the last product will be the solid content.

EXAMPLE.

If the height of a coal-pit, at the chimney, be 9 feet, and the diameter at the bottom be 24 feet, how many cords of wood does it contain, allowing nothing for the chimney?

24÷2=12-semidiam.

And 432+81×9×,5236

12x12x3=432. 9×9-81.

=18,886+ cords. Ans.

128 solid feet in a cord.

*If the diameter of a sphere be i, its solidity will be ,5236; and if its circumference be 1, its solidity will be ,016887.

SECTION III.

OF CASK GAUGING.

Definition.-Gauging is the finding of the content of any Cask, Box, Tub, or other Vessel.

Among the many different rules for gauging, the following is as exact as any.

RULE.-Take the diameter at the bung and head, and length of the cask; subtract the head diameter from the bung diameter, and note the difference.

If the staves of the cask be much curved or bulging between the bung and head, multiply the difference of diameters by,7; if not quite so much curved, by,65; if they bulge yet less, by ,6; and if they are almost or quite straight, by ,55-and add the product to the head diameter; the sum will be a mean diameter.

Square the mean diameter, thus found, and multiply the square by the length; divide the product by 359 for ale or beer gallons, and by 294 for wine gallons.

NOTE.-1. To measure the length of the cask, take the length of the stave; then take the depth of the chimes, which, with the thickness of the heads, (that are 1 inch, 1 inch, or 2 inches, according to the size of the cask) being subtracted from the length of the stave, leaves the length within.

2. In taking the bung diameter observe by moving the rod backward and forward whether the stave opposite to the bung, be thicker or thiner than the rest, and if it be, make allowance accordingly.

EXAMPLE.

How many ale and wine gallons will a cask contain, whose bung diameter is 30 inches, head diameter 25 inches and length 40 inches?

30-25-5. 5x,7=3,5 25+3,5=28,5 mean

28,5×28,5×40

diam.

--=90,5 ale gal. Ans.

359

28,5×28,5×40

-=110,51+wine gal. Ans.

Or, by the sliding rule. On D. is 18,94—the gaugepoint for ale or beer gallons, marked A. G: and 17,14— the guage-point for wine galons, marked W. G. Set the gauge-point to the length of the cask on C. and against the mean diameter, on D. you will have the answer in ale or wine gallons, accordingly as which gauge-point you make use of.*

CASE 2.-To gauge round tubs, &c.

RULE.-Multiply one diameter by the other, and to that product add one third of the square of their difference; multiply this sum by the length, and divide as before for beer or wine.

EXAMPLES.

What is the content, in beer and wine gallons, of a round tub, whose diameter at the top, within, is 40 inches, and at the bottom 34 inches, and the perpendicular height 36 inches?

CASE 3.-To gauge a square vessel.

RULE.-Multiply the length by the breadth, and that product by the depth; then divide by 282 for beer or ale, (the inches in a beer or ale gallon,)and by 231 for wine, &c., (the inches contained in a wine gallon,) and the quotient will be the answer.

* A rule which has been given as generally more exact, is this; multiply the product of the square of the mean diameter and the length, by 34, and point off four places from the right of the product; the figures on the left of the point will be the gallons, and those on the right decimal parts of a gallon, in wine or cider. Let the dimension be taken exactly in inches and tenths. Take the preceding Example in Case 1.

30+25
2

=27,5 mean diam.

and 27,5×27,5x40x34-1028500,00-102-85 gallons of wine or cider; which is 76 gal. less than by the other.

EXAMPLE,

If a square vessel he 80 inches in length, 60 in breadth, and 40 inches deep, what is its content in beer and wine gallons?

NOTE. The content of any vessel, in feet, gallons, and bushels, may be thus found: Measure the inside of the vessel, according to the rule of the figure, and find the content in cubic inches; then,

1728

and the
quotient

will be the

content in

Cubic feet.
Ale or beer gal.
Wine gallons.
Bushels.

SECTION IV. To find the tonnage of a ship.

RULE.-Multiply the length of the keel by the breadth of the beam, and that product by the depth of the hold; divide the last product by 95, and the quotient is the tonnage. If double decked, half the breadth is the depth. EXAMPLE.

If a ship be 72 feet by the keel, 24 feet by the beam, and 12 feet deep, what is the tonnage ?

72×24×12÷÷95=218,2+tons. Ans.

Or, RULE 2.-Multiply the length of the keel by the breadth of the beam, and that product again by half the breadth of the beam; divide the last product by 94, and the quotient is the tonnage.*

EXAMPLE.

What is the tonnage of a ship that is 84 feet by the keel, and 28 feet by the beam? 23÷÷2=14

84×28×14÷94=350,29+tons.

Ans.

NOTE. The breadth of the beam added to two thirds the length of the keel, gives the length of a ship's main

*Rule established by Congress. For double decked vessels; length from fore part of main stem to afterpart of stern post. above upper deck; breadth at widest part, above main wales, t.a.f of which is called the depth deduct from length 3-5ths of breadth; multiply the remainder by width

mast: Therefore the length of the mainmast of the ship last mentioned, is 84x2÷3+28-84 feet.

To find the thickness, the proportion is, as 84 to 28, so is the length of a mast in feet, to its thickness in inches. Consequently the thickness of the mast whose length was just found, is 28 inches.

SECTION V-To find the weight of anchors which cables may sustain.

RULE. As the strength of cables, and consequently the weights of their anchors, are proportioned to the cubes of their peripheries; therefore, as the cube of the periphery of any cable, is to the weight of its anchor, so is the cube of the periphery of any other cable, to the weight of its anchor.

EXAMPLES.

1. If a cable of 6 inches round require an anchor of 21cwt. what would be the weight of an anchor, for a 12 inch cable?

cwt.

6×6×6 : 2,25 :: 12×12×12 : 18cwt. Ans.

2. If a 12 inch cable required an anchor of 18cwt. what must the circumference of a cable be, for an anchor of 21cwt.? cut, cwt.

18:12×12×12::2,25:216. 216=6in. Ans. SECTION VI.-From one solid's capacity to find another's. RULE. As the cube of any dimension is to its given weight, so is the cube of any like dimensions to its weight.

EXAMPLES.

1. If a ship of 300 tons' burthen be 75 feet by the keel, what is the burthen of one, 100 feet by the keel, of like form; 25ft. a qr. ?

tons.

tons. crt. H.

753: 300 :: 1003:711 2 22. Ans.

2. If a brass cannon, 11inch. diameter, weigh 1000. what will another, 20,83; inch diameter, of like metal and shape, weigh? Ans. 5942,5697.+

and the product by depth; divide by 95: the quotient is the true tonnage. For single-decked, take depth from under side of deck plank to ceiling in the hold; and proceed as before.