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3. What is the solidity of a cone in cubic inches, the diameter at the base being 15 inches, and perpendicular height 32 inches?
152 x 7854 × 32.5
=1914-4125 cubic inches.
4. In a square solid pyramid of stone 67 feet in height, and 16 feet at the base, how many cubic feet?
To find the solidity or capacity of any frustum of a cone or pyramid.
Rule.-If the base be a circle, add into one sum the two diameters, or, if a regular polygon, the breadth of one side at the top and at the base; then from the square of the sum subtract the product of these diameters or breadths; multiply the remainder by 7854, if a circle, or by the tabular area (see Table of Polygons, p. 36) and by one-third of the height, and the product is the content in equal terms of unity.
Note. Where the whole height of the cone or pyramid can be obtained, of which the given frustum forms a part, the most simple method is first to find the whole contents, then the contents extending beyond the frustum, and subtracting the less from the greater, leaves the contents of the frustum required.
Application of the rules.
1. The perpendicular height A B of the frustum of a hexagonal pyramid C D E is 7 feet, and the breadth of each side at top and base equal 3 and 2 feet; required the solid contents of the frustum in cubic feet
3.75+2·5=6·25, and 6.25 × 6·25=39′0625; then 375 × 2.5 =9.375, and 39.0625 — 9.375 = 29.6875 × 2.598 (tabular area, p. 36) 77138 × 2.5 or of the height = 192.845
2. Required the solidity of the frustum of a cone, the top diameter of which is 7 inches, the base diameter 9, and the perpendicular height 12.
7 +9.52272.25, and 7 × 9.5 66·5; then 272.25 =205.75 × 7854 = 161.576 × 4 or 3 of the height = 646.3 cubic inches.
3. A vessel in the form of an inverted cone, as A B C D, is 5 feet in diameter at the top, 4 feet at the bottom, and 6 feet in depth; required its capacity in imperial gallons.
5+4=92=81, and 5 × 4=20; hence 81-20-61 x '7854, and by 2 or of the depth =597.1427 gallons.
95-8188 cubic feet, and × 6.232
To find the solid contents of a wedge.
Rule. To twice the length of the base add the length of the edge; multiply the sum by the breadth of the base, and by the perpendicular height from the base, and one-sixth of the product is the solid
Application of the rule.
Required the solidity of a wedge in cubic inches, the base A B C D being 9 inches by 3, the edge E F, 7 inches, and the perpendicular height GE, 15.
To find the convex surface, the solidity, or the capacity of a sphere or globe.
Rule 1.-Multiply the square of the diameter by 3.1416, the product is the convex surface.
Rule 2.-Multiply the cube of the diameter by 5236, the product is the solid contents.
Rule 3.-Multiply the cube of the diameter in feet by 3.263, or in inches by 001888, the product is the capacity in imperial gallons.
Application of the rules.
1. Required the convex surface, the solidity, and the weight in cast iron, of a sphere or ball 10 inches in diameter.
10.52 x 3.1416-346 3614 square inches.
10.53 x 5236=606·132, &c. cubic inches, and
606·132 × 263 (see Table of Approximations, p. 30)=159.4 lbs.
2. A hollow or concave copper ball is required 8 inches diameter, and in weight just sufficient to sink to its centre in common water; what is the proper thickness of copper of which it must be made?
Weight of a cubic inch of water
03617 lbs. 3225
see p. 113.
4.84828 cub. in. of water to be displaced.
=15.0334 cubic inches of copper in the ball.
Then 82 x 3.1416=201·0624, and thickness of copper required.
=0747 inch, the
0747 × 16
of an inch full, or 3 lbs. copper to a square foot.
3. What diameter must I make a leaden ball, so
as to weigh 72lbs.?
5236 x 4103=21483308, and inches diameter.
To ascertain the amount of convex surface, also the solid contents, of the segment of a globe.
Rule 1.-Multiply the circumference of the globe or sphere by the height of the segment, and the product is the convex surface.
Rule 2.-To three times the square of the segment's radius add the square of its height, multiply the sum by the height, and by 5236; the product is the solid contents.
Application of the Rules.
1. Required the number of square feet in the convex surface of a sphere, the height of which is 9 feet, and the circumference of the sphere of which it is a part equal 70 feet.
70.5 x 9.5-669.75 square feet.
2. The radius A C or B C of the spherical segment A D B equal 48 inches, and the height D C equal 12 inches; required its solidity in cubic inches.
482 × 3=6912, and 122=144; then 6912 + 144 × 12 x 5236=44334.75
To find the convex surface and solidity of a cylindrical ring.
Rule 1.-To the sectional diameter of the ring add the inner diameter of the circle, multiply the sum by the sectional diameter, and by 9.8696; the product is the convex surface.
Rule 2.-To the sectional diameter of the ring add the inner diameter of the circle, multiply the sum by the square of the sectional diameter, and by 2.4674; the product is the solid contents.
Application of the Rules.
The inner diameter A B of the cylindric ring c d e f equal 18 feet, and the sectional diameter ca or Be equal 9 inches; required the convex surface and solidity of the ring.
18 feet x 12=216 inches, and 2169 × 9 × 9.8696
216+9 x 92 x 2.4674-44968-365 cubic inches.
In the formation of a hoop or ring of wrought iron, it is found in practice, that in bending the iron the side or edge which forms the interior diameter of the hoop is upset or shortened, while at the same time the exterior diameter is drawn or lengthened; therefore, the proper diameter by which to determine the length of the iron in an unbent state, is the distance from centre to centre of the iron of which the hoop is composed: hence the rule to determine the length of the iron. If it is the interior diameter of the hoop that is given, add the thickness of the iron; but if the exterior diameter, subtract from the given diameter the thickness of the iron, multiply the sum or remainder by 3.1416, and the product is the length of the iron, in equal terms of unity.
Supposing the interior diameter of a hoop to be 32 inches, and the thickness of the iron 11, what must be the proper length of the iron, independent of any allowance for shutting?
32+1.25 33.25 × 3·1416=104.458 inches.
But the same is obtained simply by inspection in the Table of Circumferences.
Thus, 33-25=2 ft. 91 in., opposite to which is 8 ft. 8 inches.