Operation. The Bung Diameter 31.5 lefs 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 fee the whole Method made ufe of in Practice by the Officers of the Excife Duty, for finding the Contents of all Kinds of Veffels, he may find an ample Account in the Young Gauger's beft Inftructor, written by the Author of this Work. In which Work is inferted an eafy Method of finding the Content of any Cafk, without confidering the bending of the Staves, or regarding what Variety it belongs to.

THE

THE

FIVE REGULAR BODIES;

COMMONLY CALLED THE

PLATONIC SOLIDS.

Superficies are fimilar and equal, and which can be fo inclofed within a Sphere, that each Angle fhall 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, Dodecaëdron, and Icofaëdron.

Of the Tetraedron.

Def. A Tetraedron is a Body contained under four equal and equilateral plane Triangles; confequently, fuch a Solid is a Pyramid ftanding on an equilateral triangular Bafe; 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 Pafteboard,

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 Tetraedron; and this I would advife 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 Tetraedron.

Rule.

of the

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

Example.

Suppofe ABCDE to represent a Tetraedron, each of whofe 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 Bafe, 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; confequently the Solidity is equal to three Dimenfions 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 Pafteboard fimilar to this here inserted, 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 unneceffary to infert it here, unlefs to preferve the Propriety of reprefenting the Five Platonic Solids in a regular Order.

Problem 2.

To find the Solidity of an Hexaëdron.

Rule.

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

Suppofe the Side

Example.

=

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.