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Operation. The Bung Diameter 31.5 less the Head Diameter 24,5 is equal to 7; which multiplied by the Numbers in the above Rule, according as the Cask is more or less arching, and added to the Head Diameter, the leveral mean Diameters will be for the

i Variety 2 Variety 3 Variety 4 Variety

29.4 29.05 28.7

and Content of
each Calk will
be found to be


Ale 96.35 Gallons, 94.03


But if the Young Geometrician wishes to see the whole Method made use of in Practice by the Officers of the Excise Duty, for finding the Contents of all kinds of Vessels, he may find an ample Account in the Young Gauger's besi Inftruttor, written by the Author of this work. In which Work is inserted an easy Method of finding the Content of any Cask, without considering the bending of the Staves, or regarding what Variety it belongs to.




Superficies are similar and equal, and which can be so inclosed within a Sphere, that each Angle thall touch the internal Surface of that Sphere, are called Regular Bodies ; of which Bodies there are in Nature no more than five, viz. the Tetraëdron, Hexaëdron, OEtaëdron, Dodecaëdrong and Icofaëdron.

Of the Tetraëdron. Def. A Tetraëdron is a Body contained under four equal and equilateral plane Triangles; confequently, such a Solid is a Pyramid standing on an equilateral triangular Base; and its superficies is equal to four Times the Area of the Base.

To conceive a more perfect Idea of the Tetraëdron, the Learner may draw a Figure like this upon Pafteboard,

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or any other pliable Matter; then by cutting the Lines half through, turning up the Parts, and gluing them together, they will form a complete Tetraëdron; and this I would advise him to do, as he may by that Means more readily perceive the Reason of the following Operations.

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Multiply, the Area of one of the Triangles by of the Perpendicular Height of the Figure, and the Product will be the Solid Content,

Crample. Suppose A B C D E to represent a Tetraëdron, each of whose Sides, as AB, BC, and CA, are 6 Inches; the Height D E of one of the Triangles 5.196 Inches; and the Perpendicular Height of the Figure 4.899 Inches, what is its Solid Content?

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Operation. 5.196, the Height of one Triangle, multiplied by 3, Half the side of the Base, gives 15.588, the Area of one Triangle, which multiplied by 1.633, one-third of the Perpendicular Height of the Tetraedron, gives 25.455204

İnches, the Solid Content required. For the Superficial Content, multiply 15:588, the Area of one Triangle, by 4, the Number of Triangles, and it makes 62.352 Inches, the Superficial Content.

Of the Hexaëdron. Def. An Hexaëdron, or Cube, hath equal Length, Breadth, and Depth, like a Die, and is contained under fix Square Planes, consequently the Solidity is equal to three Dimensions multiplied by each other, and the Superficies equal to fix Times the Area of the Base, or one of its Sides.

A Figure drawn upon Pasteboard similar to this here inserted, having the Lines cut half through, folded up, and

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glued together, will form the true Figure of an Hexaëdron Problem 2.

or Cube. *

* An Hexaëdron being the fame Figure as the Cube, the Measure of which having been considered before in Stereometry, makes it unnecessary to insert it here, unless to preserve the Propriety of representing the Five Platonic Solids in a regular Order.

To find the Solidity of an Hexaëdron.


The Hexoëdron being only a Cube, is measured by mul. tiplying the Side into itself, and that Product again by the Side, and this last Product will be the Solidity required.

Example. Suppose the Side AB=BC=CD, &c. of an Hexaë dr on be 6 Inches, what is the Solid Content?

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Operation. The Side 6 multiplied by 6, gives 36, which multiplied again by 6 gives 216 Inches, the Solidity required.

For the Superficial Content, multiply 36, the Area of one of the Sides, by 6, the Number of Sides, gives 216 Inches, the Superficies fought.

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