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Problem 8.
To find the Solidity of an Hyperbolic Conoid.

Def. An Hyperbólic Conoid is a Solid made by the Revolution of a Semi-hyperbola about its Axis. *

Rule. To the Square of the Radius of the Base, add the Square of the Diameter in the Middle between the Top and Bottom; this Sum multiplied by the Height, and the Product multiplied again by .5236, will give the Solid Content.

Erample. Suppose ABCDEF be a Hyperbolic Conoid; the Semi-diameter AF of whose Base A E is 52 Inches; the Diameter in the Middle BD 68 Inches, and the Height CF 50 Inches, what is its Solid Content?

B

A

Operation. OAF 52 = 2704 + OBD 4524. = 7328 x CF 50 = 366400 X .5236 = 191847.04 Inches, the Solid Content required.

* An Hyperbolic Conoid is a Solid whose Sides are straiter than a Parabolic Conoid, yet more curved than a Cone.

Problem 9. To find the Solidity of the Fruflum of an Hyperbolic Conoid.

Rule.

To the Sum of the Squares of the Semi-diameters of the Bottom and Top of the Fruftum, add the Square of the whole Diameter in the Middle; this Sum being multiplied by the Height, and that Product again by :5236, will give the Solid Content,

#rample. Suppose ABCD be the Frustum of an Hyperbolic Conoid; the Semi-diameter A T of the Bottom measures 16 Inches; the Semi-diameter BH of the Top 12 Inches the Middle Diameter m d 28.17 Inches, and the Height HT 20 Inches, what is its Solid Content?

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Operation. O AT 256 + OBH 144 + om d 793.5489 = 1193.5489 x HT 20 = 23870.9780 x.5236 = 12498.84408080 Inches, the Solidity required.

Note. As many Houshold Utensils are in the Shape of some of the foregoing Figures, as, for Example, Tuns and Tubs in Form of Frustums of Cones or Conoids ; Furnaces and Coppers in Form of Parabolic or Hyperbolic Co. noids; * Casks in Form of the Middle Zones of Spheroids, Parabolic Spindles, Double Frustums of Parabolic Conoids, and Double Fruftums of a Cone; the Quantity of Liquor contained in each may be easily ascertained by dividing (as before in Planometry and Stereometry) the Solid Content in Inches by

282 for Ale Gallons.

231 for Wine Gallons. 2150.42 for Corn Bushels.

With respect to Casks, it may be difficult, on Account of the different Bending of the Staves, to ascertain exactly the Form to which they belong; for though the Dimenfions of several Casks may be exactly the fame, yet their Contents will be very different, as is clear from a sight of the following Figure.

* The rising Crowns of Stills are Segments of spberes; the remaining Part generally the Frustum of a Parabolic Conoid; Bowl and Bafons are generally the Segments of Spheres, and measured accordingly,

Suppofe

Suppose ABCDEF to represent a Cask; then, it is evident, that if the outer curved Lines A B C and Ó EF are the Boundaries or Staves of the Cask, it must of Course hold more than if the inner and straiter Lines were the, Bounds and Staves of it, yet the Dimensions of the Bung Diameter B E, and Head Diameters A F and CD, and the Length LH, are the same in all the Casks.

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If the Staves of the Cask are very much curved or arching, as the outer Line in the foregoing Figure, it is supposed to be in the Form of the Middle Zone of a Spheroid, and its Content may be found by Problem 3d.

If the Staves are not quite so much curved or arching, as represented in the second Line in the Figure, it is taken for the Middle Zone of a Parabolic Spindle, and is measured by Problem 7th.

When the Staves are but little curved or arching, as the third Line in the Figure, it is supposed to be in the Form of the lower Fruslums of two equal Parabolic Conoids joined together upon one common Base in the Middle, and its Content may be found by Problem gth.

If the Staves are quite strait from Bung to Head, as the inner Lines in the Figure, it is then considered as the lower Fruftums of two equal Cones joined together upon one common Base, and its Content may be found by Problem 8th. in Stereometry.

Note. Carks made in the first Form hold the most; and those of the last Form hold the least of any other Kinds,

But since we can only at last guess, as it were, at the Variety or Form which the Caik belongs to, the eafielt and best Way of finding its Content is to be preferred in Practice, which is to find such a mean Diameter between the Bung and Head Diameters as will reduce the Cask to a Cylinder equal to it, which may be done by the following

Rule.

Multiply the Difference between the Bung and Head Diameters by :7, or by .65, or by .6, or by:55, according as the Staves are more or less arching; add the Product to the Head Diameter, and that Sum will be a mean Diameter, i. e. it will be the Diameter of a Cylinder, whose Length and Content are equal, as near as can be, to that of the Cask.

Example.

Suppose a Cask whose Bung Diameter is 31.5 Inches; the Head Diameter 24.5 Inches, and its Length 42 Inches, what is its Content in Ale Gallons ?

Operation.

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