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height. Then,5236h × 3rrhh the solidity of the seg ment ABC.

B

Fig. 23.

EXAMPLE.

(26) What is the solidity of a segment 4 feet high, cut from a globe 18 feet diameter?"

PROBLEM XXVI.

To find the solid content of a spheroid.

Fig. 24. Af

RULE I.

Multiply continually together the fixed axis, the square of the revolving axis, and the number,5236 (being of 3,14159 nearly; and the last product will be the content required; that is, if p=3,14159, &c. t=the transverse, and c=the conjugate axis of the generating ellipse;

Then pttc the oblate.

Andpice the oblong spheroid.

RULE II.

Multiply the area of the generating ellipse by of the revolving axis, and the product will be the content of the spheroid.

Let A the area of the ellipse: then from the former rule Athe oblate,

And cAthe oblong spheroid.

EXAMPLE.

(27) What is the solid content of a spheroid, whose diameter of the greatest circle is 33 inches, and the length 55 inches?

PROBLEM XXVII.

To find the solidity of a parabolic spindle.

Fig. 25. A

B

RULES.

8

1. The square of the diameter (CD) of the greatest circle multiplied by ,41888 (being of ,7854), and that product again by its length (AB) will be the solidity.-Or

2. Multiply the area of the greatest circle, or middle section by the length; and of the product will be the

content.

8

That is, if A B➡the length or axis, DC=the greatest diameter, or double the abscissa of the generating parabola ACB, and n-,785398, or,7854. Then x DC2X AB= the whole solid ADBCA.

EXAMPLE.

(28) What is the solidity of a parabolic spindle, whose greatest diameter is 36, and its length 99 inches?

PROBLEM XXVIII.

To measure TIMBER.

A square piece of timber, equally thick at both ends, is a prism; a round piece, equally thick at both ends, is

a cylinder; a square piece, that tapers regularly, is the frustum of a pyramid; and a round piece, that tapers regularly, is the frustum of a cone; and the contents of these solids may be exactly computed by their respective rules.

But because the mensuration of tapering timber by the exact rules is troublesome, an approximation has taken place; and the contents of such trees are generally computed by the following

RULE.

Multiply the square of the quarter girt (or of the circumference) in inches, by the length in feet; divide the product by 144; and the quotient will be the content in feet.

But to find the content more near the truth, observe the following

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1. The girt of a piece of timber is its compass or circumference at the middle, of which is commonly taken for the side of a square, equal to the area of the section there.

2. Trees of irregular growth must be measured in parts or pieces, as above directed.

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3. Allowance must be made for the thickness of bark (if on the tree) in oak or, in other wood not so much.

4. When the timber is to be reduced to loads,

Divide the feet in

Srough,

{hewn,} by {10} gives the loads;

50

as 40 feet make one load of rough timber, and 50 one of bewn.

EXAMPLES.

*

(29) What is the content of a tree, whose girt is 42 inches, and length 16 feet an

(30) What is the content of a tree, whose compass is 6+ inches, and the length 30 feet?

(31) How many loads of timber are there in a hewn tree, whose breadth is 42 inches, depth 30 inches, and length 40 feet?

GAUGING.

PROBLEM XXIX.

To find the area of any triangular tun, back, cooler, or circular and elliptical superficies, in ale gallons, &c.

RULES.

1. Find the area in inches by the different Problems in Sect. LXXV. and the solidity by Sect. LXXVI.; then,

Divide by $282

231

for

268,8

Ale,
Wine,

Corn,

and the quotient will be the area in gallons.-Or, 2. If the square of the diameter of any circle;

359,05

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ale gallon,
wine gallon,
corn gallon,

and the quotient will be the area in their respective gallons. For as,785398:1:: 282: the square of the diameter of the circle whose area is 282 cubic inches, viz. one ale gallon: and from this proportion arises the preceding divisions:

282 Viz. 231

268,8

359,05

785398: - 294,12

342,24

Or, these divisors may be turned into multiplicators, by dividing unity or 1, or rather by dividing the area in inches, of that circle whose diameter is 1.

That is 785398 by 282, &c.

£282=,002785

282,002785

Thus,,785398÷231=,00399

268,8002922

The product will be the area in gallons of the same

name.

EXAMPLES.

(32) Suppose the length of a brewer's tun, back, or cooler, be 16 feet 6 inches, and its breadth 7 feet 4 inches, what will be the area in ale or beer gallons, &c. ?

(33) The length of the base of a triangular cooler is 94 inches, and its perpendicular breadth is 58 feet 6 inches. Required its area in ale gallons.

(34) Suppose the longest diameter of a brewer's vessel be 84,5 inches, and the shortest diameter to be 50 inches; what will be the area in ale gallons?

(35) Suppose a tun in the form of the frustum of a pyramid, whose bases are equilateral triangles: let the side of the top be 64 inches, the side of the bottom be 98,6 inches, and its height or depth 36 inches; what is the content of that tun in ale gallons, &c.?

(36) If the diameter of the base of a regular cone be 60,5 inches, and the perpendicular height be 42,8 inches, what will be the content in ale gallons, &c. ?

(37) Suppose the diameter of a frustum of a cone be 84 inches at the top, and the diameter at the bottom be 62 inches, and the height 42 inches: required the content in ale gallons.

The bung diameter E F, head diameter CD, and length of the cask A B (withinside) being given; to find the content of a cylinder nearly equal to it.

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1. To twice the area of the circle at the bung, add the area of the circle of the head; multiply the sum by onethird of the length of the cask; the product is the content

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