Problem 3. To find the Solidity of the Middle Zone of a Spheroid. Def. The Zone of a Spheroid is the Part remaining when the two Ends are cut off by Planes parallel to each other; if the Planes are equally distant from the Middle, it is called the Middle Zone of the Spheroid. Rule. To twice the Square of the Diameter in the Middle, add the Square of the Diameter of either End; multiply this Sum by the Length of the Zone, and that Product multiply again by .2618, a Decimal, and the last Product will give the Solid Content. Example. Suppose ABCDEF be the Fruftum, or Middle Zone of a Spheroid; the Middle Diameter BE is 50 Inches; the Diameter of either End AF or CD 40 Inches; and the Length LN 18 Inches; what is its Solid Content? Operation. 2BE 5000 + CD or AF = 1600 = 6600 × LN 18 = 118800, which × .2618 gives 31101.8400 Inches, the Solidity fought. Problem 4. To find the Solidity of a Parabolic Conoid. Def. A Parabolic Conoid is a Solid, formed by the Revolution of a Semi-Parabola about its Axis. Rule. Multiply the Area of the Base, by Half the Height, and the Product will be the Solidity required. * Example. Suppose ABCD be a Parabolic Conoid, whose Diameter of its Base A C is 40 Inches, and Height BD is 30 Inches, what is its Solid Content? Operation. A C 40 × 40 = 1600 ×.78541256.6400, the Area of the Base, which × 15 half BD = 18849.6000 Inches, the Solidity required. * Every Parabolic Conoid is equal to its circumfcribing Cylinder. Problem 5. To find the Solidity of a Fruftum of a Parabolic Conoid. Rule. Multiply the Sum of of the Squares of the Bottom and Top Diameters by the Frustum's Height, and that Product multiply again by . 3927, * a Decimal, and it will give the Solidity required. Example. Suppose ABCD be the Fruftum of a Parabolic Conoid, the greater Diameter A D being 58 Inches, the lesser Diameter BC 30 Inches, and the Height g h 18 Inches, what is the Solid Content? Operation. AD 3364+ BC 900 = 4264 × gb 18 = 76752 × 3927 = 30140.5104 Inches, the Solidity required. * 3927 is the Half of 7854, the Area of a Circle whose Diameter is 1, Problem 6. To find the Solidity of a Parabolic Spindle. Def. A Parabolic Spindle is a Solid generated by the Revolution of a Parabola about its greatest Ordinate or Bafe. Rule. Multiply the Area of the Circle in the Middle by the Length, which Product multiplied again by 8, and that Product divided by 15, gives in the Quotient the Solidity required. * Example. Suppose ABCD be a Parabolic Spindle, whose Diameter in the Middle AC is 34 Inches, and the Length BD is 60 Inches, what is its Solid Content? Operation. AC = 34 × 34 = 1156 × 7854 = 907.9224, the Area of the Circle in the Middle, which × BD = 60 = 54475.3440, which × 8 = 435802.7520 15=29053.5168 Inches, the Solidity fought. * Every Parabolic Spindle is equal to linder; or, multiply the Square of the Diameter in the Middle by the to s of its circumscribing CyLength, and that Product again by .418879 of 7854. Problem 7. To find the Solidity of the Middle Zone of a Parabolic Spindle. Rule. To twice the Square of the Middle Diameter add the Square of the Diameter of the End; from this Sum fubtract .4 (Tenths) of the Square of the Difference of the two Diameters; multiply the Remainder by the Length, and that Product divided by 3.8196 will give the Solidity of the Zone required. Example. Suppose ABCDEF be the Middle Zone of a Parabolic Spindle; the Diameter of which in the Middle BE is 36 Inches; the Diameter at the End CD or AF is 20 Inches; and the Length In 36 Inches, what is its Solid Content? Operation. BE 36 × 36×2+CD 20×20=2992; and BE 36-CD 20 = 16, the Difference of the two Diameters. Again, 16 × 16 = 256, 4-tenths of which is = 102.4. Then, 2992 - 102.4 × 1 n 36 = 104025.6, which 3.8196 gives 27234.68, &c. Inches, the Solidity fought. |