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SECTION III.

OF SOLIDS PROPERLY SO CALLED.

PROBLEM I.

To determine the Content of a Cube.

RULE.

By the Pen.

Multiply the side of the cube by itself, and the product again by the side. Then divide the last product by the number of cubic inches in the proposed integer.

By the Sliding Rule.

Set the square gage point, for the proposed integer on D, to the number expressing the side of the cube on C, then against the same number on D, will be found the content on C.

EXAMPLE.

Let ABCDEFG be a cistern in the form of a cube, whereof the side is 100 inches; the content is required in wine gallons, ale gallons, and malt bushels.

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Multiply by 100 side

Product 10000 square of the side

Multiply by 100 side

Product 1000000 cubic inches in the cube.

And,

W.G. divisor 231)1000000 (4329.00 wine gallons.

A. G. divisor 282)1000000 (3546.10 ale gallons.

M. B. divisor 2150)1000000 ( 465-11 malt bushels.

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Sq.G.P. 16.79 to 100 against 100 are 3546-1 A. G.

46.37

465.1 M. B.

PROBLEM II.

To determine the Content of a Parallelopiped on.

RULE.

By the Pen.

Multiply the length by the breadth, and the product by the depth: then divide the last product by the number of cubic inches in the proposed integer.

By the Sliding Rule.

Find a mean proportional between the length and breadth; then set the square gage point on D, to the depth on C, and opposite to the mean proportional on D, will be the content on C.

EXAMPLE.

Let DEFGHIK be a box in the form of a parallelopipedon, the length DE being 80 inches; the breadth, EG, 20 inches; and the depth, EK, 30 inches: the content is required in wine gallons, ale gallons, and

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SOLUTION BY THE PEN.

80 length

Multiply by 20 breadth

Product 1600 area of the base.
Multiply by 30 breadth

W.G. divisor 231) 48000 (207-79 wine gallons.

A. G. divisor 282) 48000 (170.21 ale gallons.

M. B. divisor 2150) 48000 (22.32 malt bushels.

BY THE SLIDING RULE.

Set 80 on C, to 80 on D, and against 20 on C, are 40 on D, for the mean proportional between the length and breadth of the box.

[blocks in formation]

Sq. G.P. 16.79 to 30 against 40 are 170-2 A. G.

46.37

PROBLEM III.

22.3 M.B.J

To determine the Content of a Cylinder.

RULE.

1

By the Pen.

Find the area of the base, and multiply it by the depth: then divide the product (if in cubic inches)

by the number of cubic inches in the proposed integer.

By the Sliding Rule.

Set the circular gage point on D, to the altitude or depth on C, and against the diameter on D, will be the content on C.

EXAMPLE.

Let KLMNOPQ be a cylindrical vessel, the diameter KM, being 38 inches, and the depth NP, 15-16 inches; the content is required in wine gallons, ale gallons, and malt bushels.

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Product 1444 square of the diameter
Multiply by 7854 constant decimal

Product 1134-117

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