Problem 8. Def. An Hyperbolic 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. Crample. Suppose ABCDEF be a Hyperbolic Conoid; the Semi-diameter A F of whose_Base A E is 52 Inches; the Diameter in the Middle B D 68 Inches, and the Height CF 50 Inches, what is its Solid Content? Operation. OAF 52 = 2704 + o BD 4524 = 7328 xCF 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. Crample. Suppose ABCD be the Frustum of an Hyperbolic Conoid; the Semi-diameter AT 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 ? Operation. O AT 256 + OBH 144 + Omd 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; Fure. naces and Coppers in Form of Parabolic or Hyperbolic Conoids; * Casks in Form of the Middle Zones of Spheroids, Parabolic Spindles, Double Fruftums 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 same, 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 ; Bowls and Bafons are generally the Segments of Spheres, and measured accordingly. Suppose ABCDEF to represent a Calk; then, it is evident, that if the outer curved Lines A B C and D 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 BE, and Head Diameters AF and CD, and the Length LH, are the same in all the Casks. If the Staves of the Cark 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 Frufiums 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 Sth. 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 easiest 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. |