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 feveral mean Diameters will be for the

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 best Instructor, written by the Author of this Work. In which Work is inferted an easy Method of finding the Content of any Cask, without confidering the bending of the Staves, or regarding what Variety it belongs to.

THE THE

FIVE REGULAR BODIES;

A

COMMONLY CALLED THE

PLATONIC

SOLIDS.

LL those Solid Bodies whose Sides are equal, and whose Superficies are fimilar and equal, and which can be fo inclosed within a Sphere, that each Angle shall touch the internal Surface of that Sphere, are called Regular Bodies; of which Bodies there are in Nature no more than five, viz. the Tetraedron, Hexaëdron, Octaëdron, Dodecaedron, and Icofaëdron.

Of the Tetraedron.

Def. A Tetraëdron is a Body contained under four equal and equilateral plane Triangles; consequently, 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 Tetraedron, the Learner may draw a Figure like this upon Pasteboard,

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.

Problem 1.

To find the Solidity of a Tetraëdron.

Rule.

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

Example.

Suppose ABCDE to represent a Tetraedron, each of whose Sides, as AB, BC, and CA, are 6 Inches; the Height DE of one of the Triangles 5.196 Inches; and the Perpendicular Height of the Figure 4.899 Inches, what is its Solid Content?

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 Inches, 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 inferted, having the Lines cut half through, folded up, and

glued together, will form the true Figure of an Hexaëdron or Cube. *

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

Problem 2.

To find the Solidity of an Hexaëdron.

Rule.

The Hexoedron being only a Cube, is measured by multiplying 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

Hexaëdron be 6 Inches, what is the Solid Content ?

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.