32147.025 6 =5357-8375 cubic inches 17283.1006 feet. Answer. 2. If the base of a wedge be 27 inches and 8 inches, the edge 36 inches, and the height 3.5 feet, what is the solidity? Ans. 2.9166 cubic feet. RULE. PROBLEM 20. To find the solidity of a prismoid. To the sum of the areas of the two ends add four times the area of the section parallel to, and equally distant from both ends, and this last sum multiplied by of the height will give the solidity. 1. What is the solidity of a rectangular prismoid, the length and breadth of one end being 12 and 8 inches, and the corresponding sides of the other 8 and 6 inches, and the perpendicular height 60 inches? = = Thus, 12 × 8 96; 8 × 6 48 +96 = 144, sum of the areas of the two ends. Then 12+8=20 ÷ 2 = 10, length of the middle rectangle. 8+6=14 2 = Hence 4 x 10 x 7 of the middle rectangle. 144 +280 x 60 = Prismoid. 7, breadth of the middle rectangle. 424 x 10 = 4240 cubic inches. And 42401728 2.4537 feet. Ans. 2. What is the solidity of a stick of hewn timber whose ends are respectively 30 by 27 inches, and 24 by 18 inches, and whose length is 48 feet? Ans. 204 feet. PROBLEM 21. To find the convex superficies of a cylindric ring. RULE. To the thickness of the ring add the inner diameter, and this sum being multiplied by the thickness, and the product again by 9:8696, (or the square of 3-1416,) will give the superficies required. 3. The thickness of a cylindric ring is 2 inches, and the inner diameter 12 inches; what is the convex superficies? PROBLEM 22. Ans. 276-3488 inches. To find the solidity of a cylindric ring. RULE. To the thickness of the ring add the inner diameter, and this sum being multiplied by the square of half the thickness, and the product again by 9-8696, will give the solidity. 1. What is the solidity of an anchor ring whose inner diameter is 8 inches, and thickness of metal 3 inches? Thus, 8 +3 = 11; 1·522.25 × 11 244.2726 cubic inches. Ans. 1= = 24.75 9.8696 = 2. The inner diameter of a cylindric ring is 12 inches, and its thickness 4 inches; what is its solidity? Ans. 631-6544 inches. 3. Required the solidity of a cylindric ring whose inner diameter is 12 inches, and thickness 5 inches. Ans. 1048-645 inches. 4. What is the solidity of an anchor ring whose inner diameter is 9 inches, and the thickness of metal 3 inches? PROBLEM 23. Ans. 266-4792 inches. The solidity and thickness of a cylindric ring being given, to find the inner diameter. RULE. Divide the solidity of 9.8696, and that quotient by the square of half the thickness; from which subtract the thickness, and the remainder will be the inner diameter of the ring. 1. The thickness of a cylindric ring is 4 inches, and its solidity 789-568 solid inches; what is its inner diameter ? Thus, 789-568 ÷ 9.8696 inches, diameter. = 80÷22= (4) = =20 4 = 16 2. What is the inner diameter of a cylindric ring whose solidity is 1 solid foot, and thickness 4 inches? 3. What is the inner diameter of a cylindric ring whose solidity is 244-2726 inches, and thickness 3 inches. PROBLEM 24. Ans. 30.77 in. To find the convex surface of a sphere. Ans. 8 in. RULE.-Multiply the diameter of the sphere by its circumference, and the product will be the convex superficies required. The curve surface of any zone or segment will also be found by multiplying its height by the whole circumference of the sphere. 1. What is the convex surface of a sphere or globe A D B C, whose diameter A B is 16 inches? Thus, 3-1416 x 16- =50.2656 x 16 804-2496 square inches, the surface A required. = 2. What is the convex superficies of a sphere whose diameter is 13 feet, and the circumference 4.1888 feet? Thus, 11 = 1/32 4, 4 16.7552+ 3 = 5.58506 feet. Ans. Sphere, or Globe. 3. If the diameter or axis of the earth be 7957 miles, what is the whole surface, supposing it to be a perfect sphere? Thus, 7957-75 x 3.1416 = 25000-0674, the circumference; 7957.75 x 25000-0674 = 198944286-35235 sq. miles. Ans. 4. What is the area of the convex surface of a globe whose diameter is 4 feet? Ans. 50-2656 square feet. 5. What is the convex surface of a sphere whose diameter is 6 feet? Ans. 113.0976 feet. PROBLEM 25. To find the solidity of a sphere or globe. RULE.-Multiply the cube of the diameter by 5236, and the product will be the solidity. 1. What is the solidity of a sphere whose diameter is 11 feet? 43 64 Thus, 11 = 3 ×·5236 = × 523633·3104 ÷ 27 = 27 1.2411 feet. Ans. 2. What is the solidity of the earth, supposing it to be perfectly spherical, and the diameter 7957-75 miles? Thus, 7957-753 ×·5236 = 63325785-0625 × 7957-75 x 5236 =503930766081-109375 × 5236 =263858149120-06886875. Answer. 3. What is the solidity of a globe whose diameter is 3 feet 4 inches? Ans. 19-3926 cubic feet. 4. What is the solidity of a globe whose diameter is 17 inches? Ans. 1-4493 feet. 5. What is the solidity of a sphere whose diameter is 6? Ans. 113-0976. 6. What is the solidity of a sphere whose diameter is 10? Ans. 4188.8. PROBLEM 26. The convex surface of a globe being given, to find its diameter. RULE.-Multiply the given area by 31831, and the square root of the product will be the diameter. 1. What is the diameter of that globe, the area of whose convex surface is 14 square feet? Thus, 14 x 318314-456342-1110 feet, the diameter required. 2. The convex surface of a sphere is 1 square rood; required the diameter. Ans. 3-5682 rods. 3. The expense of gilding a ball, at D. 1.80 per square foot, is 34 dollars; required the diameter. Ans. 2-452 feet. PROBLEM 27. The solidity of a globe being given, to find the diameter. RULE. Divide the solidity by 5236, and extract the cube root of the quotient. 1. The solidity of a globe is 2000 solid inches; what is the diameter ? Thus, 20005236 = √3819·7097 15.631 inches. Ans. 2. The solidity of a globe is 10 solid feet; required the diAns. 2.67 feet. ameter. PROBLEM 28. To find the convex surface of a spherical zone. RULE.-Multiply the height of the zone by the circumference of a great circle of the sphere, and the product will be the convex surface. 1. What is the convex surface of the zone A B D, the height B E being 9 inches, and the diameter of the sphere 42 inches? 3. The diameter of a sphere is 25 feet, and height of the zone 4 feet; what is the surface of the zone? Ans. 314.16 square feet. PROBLEM 29. To find the solidity of a spherical segment with one base. RULE. TO 3 times the square of the radius of the base add the square of the height. Multiply this sum by the height, and the product by 5236; the result will be the solidity of the segment. 1. What is the solidity of the segment A B D, the height B E being 4 feet, and the diameter AD of the base being 14 Abrup feet? Thus, 7 x7 49 x 3 = 16) = 163 x 4 x 5236 = 341-3872 solid feet, which is the solidity of the segment A B D. 2. What is the solidity of a spherical segment, the diameter of its base being 17-23368, and height 4.5? Ans. 572.5566. 3. What is the solidity of a segment, when the diameter of the sphere is 20, and the altitude of the segment 9 feet? Ans. 1781.2872 cubic feet. |