To the square of the diameter of the greater end, add the square of the diameter of the lesser end ; multiply the sum by 3927, and the product by the height, will give the solidity required. Ex. What is the solidity of a parabolic frustrum, the diameter of the greater end being 60 feet, the lesser end 48 feet, and the distance of the ends 18 feet? 60° +48°=5904 the sum of the squares of the ends. Then 5904 x 3927 x 18=41733 0144 the solidity required. If a circle is carried round a right line as an axis, and in the same plane with the circle, either touching the axis, or at a given distance from it, it will generate a solid, called an annulus, or cylmdric ring. The diameter of a generating circle is called the thickness of the ring Twice the distance of the generating circle, from the axis of rotation, is called the inner diameter. Prob. 27. To find the solidity of an annulus, or ring, whose thickness and inner diameter are known. Method 1. To the thickness of the annulus, add the inner diameter; multiply the sum by the square of the thickness, and the product by 2-4674, will give the solidity sought. Method 2. Multiply the circumference round the middle of the annulus, or that circle generated by the centre of the generating circle, by the area of the generating circle, and the product will give the solidity. Nole. This last method will give the solidity of any part of an annulus, or ring, comprehended between any two planes passing through the fixed axis. Er. What is the solidity of an annulus, whose inner diameter is 8 inches, and the thickness of the annulus 3 inches : Then 8+3 x 39 x 2.4674=244.2726 the solidity required. Prob. 23. To find the solidity of a hollow cylinder, the exterior and interior diameters being given, and the perpendicular height. Multiply the sum of the diameters by their difference, and the product by •7854 ; then by the altitude of the cylinder, and you will have the solidity required. Er. Required the solidity of a hollow cylinder, the exterior diameter being 14 feet, the interior diameter 12 feet, and the perpendicular height 10 feet. (14+12)(14-12):7854x10=408.408 the solidity required. Prob. 24. To find the solidity of a frustrum of a hollow cone, of an equal thickness; the exterior and interior diameters at each end being given, and the perpendicular altitude. Multiply together the sum of the two interior diameters, and the difference of the diameters at either end, by •7854, and by the per. pendicular altitude; then by the said difference, and the continued product will give the solidity required. Es. Required the solidity of a frustrum of a hollow cone, the boto fom diameters being respectively 35 and 32 feet; w diameters respectively 29 and 26 feet; and the perpendicular altitude 25 feet. Nov 35-32=3 and (32+26+3) = 61-7854 x 61 x 3 x 25 = 8593 205 the solidity required. CONVEX SIDE OF SOLIDS. Prob. 25. To find the convex surface of a right cylinder, the cir. cumference and length of the cylinder being given. Multiply the circumference by the length of the cylinder, and the product will be the area. Er. What is the convex superficies of a right cylinder, whose circumference is 2ft. Gin. and the length 5. 3in. Then (5f. 3i.) x (2f. 6.)=13f. li. ii. the answer. Note.-If the diameter is given, find the circumference, and proceed as before. Prob, 26. To find the convex superficies of a right cone, the circumference and slant side being given. Multiply half the circumference by the slant side of the cone, and the product will be the area. Ex. What is the convex superficies of a right cone, the circumference of the base being 4.5ft. and the slant side 6'25ft. ? 4.5 x 625 Note.-If the diameter is given, find the circunference, and proceed as before. Prob. 27. To find the convex surface of the frustrum of a right cone, the circumferences of both ends being given, and the slant side of the cone. Multiply half the sum of the circumferences by the slaust side of the cone, and the product will be the area. Er. What is the convex superficies of a frustrum of a right cone, the circumference of the base being 4.6ft. the circumference of the top being 8.25ft. and the altitude 5.75 feet? 4:6+3:25 2 Prob. 28. To find the superficies of a sphere or globe, the greatest circumference being given. Multiply the square of the circumference by '3183, and the product will be the superficies. Er. What is the superficies of a globe, the greatest circumference being 10.6f. ? Here (106)* x3183=35704188 Prob. 29. To find the convex superficies of the segment of a sphere ór globe, the diameter of the base of the segment, aad its height, being given. To the square of the diameter of the base, add the square of twice the height; and the sum multiplied by •7854, will give the superficies. Ex. What is the convex surface of the segment of a globe, the diameter of the base being 17:25 feet, and the height 4:5 feet ? 9=81 square of twice the height. (17.25)=297:5625 square of the diameter of the base. Then (297:5625 +81) * •7854=297.3229875 the superficies required. Prob. 30. To find the convex surfaces of a spherical zone, the diameters of the ends and their distance being given. Find the diameter of the sphere by Problems 13 and 14; multiply the diameter of the sphere, and the distance of the parallel ends of the zone together, and the product by 3•1416, will give the superficies required. Er. In a spherical zone, the distance of the parallel ends being 4in. the diameter of the greater end 24in. and that of the lesser end 2010. what is the convex superficies, when the centre of the sphere is with. out the zone? The distance of the greater chord from the centre, will be found to te 3.5 inches, by Prob. 13. The radius will be found to be 25 inches, by Prob. 14, or the diameter 50 inches. Therefore, 50 x 4x3.1416=628·32 the answer. Note.- If the diameter is given, find the circumference, and proceed as before. Prob. 31. To find the superficies of an annulus, or ring, whose thickness and inner diameters are known. Method 1. To the thickness of the ring, add the inner diameter ; multiply the sum by the thickness, and the product by 9.869, will give the superficies required. Method 2. Multiply the circumference of the generating circle by the circumference round the middle of the ring, or that line generated by the centre of the generating circle, and the product will be the area. Note. This last method will give the convex superficies of any part of an annulus, or ring, comprehended between two planes passing through the first axis. Ex. What is the convex superficies of an annulus, or ripg, whose inner diameter is sin, and the thickness 3in. ? Then (3+8) x 3 x9 869=325.677 the superficies required OF SPECIFIC GRAVITY. Def. 1. The specific gravity of a body, is the relation that the weight of a magnitude of one kind of body, has to the weight of an equal magnitude of another kind. 2. In this comparison of the weight of bodies, it is convenient to consider one body as the standard or unit, to which the others are to be cumpared ; and as rain water is nearly alike in all places, it is the most convenient standard. 3. It has been found, by repeated experiments, that a cabic foot of rain water weighed 624 pounds avoirdupois ; consequently, (0323) 1728 0 03616898lb. is the weight of one cubic inch of rain water. 4. The knowledge of the specific gravities of bodies, is of great use in computing the weights of such bodies, as are too beavy or too unshapely to have their weight discovered by other means A TABLE Showir:g the specific gravity to rain water, of metals, and other bodies ; and the weight of a cutic inch of each, in parts of a pound avoirdupois, and an ounce troy. Bodies. Sp. gra. Wt. Ib. av. Wt. oz. tr. Fine gold 19-640 0*7103587 10•359273 ... ... ... 1 888 TABLE CONTINUBD. JED. Bodies. Sy gra. Wt. Ib. av. Wt. oz. tr. 0.958489 ******** } Ivory 1.832 0·0662606 0-09479862 0·0237630 0:613 0·0221715 0.569 0 0205801 0.240 0.0186805 0.0012 0.0000434 0-949494 0-904498 0.699936 0.661959 0.606576 0:560691 0·5-4397% 0°543272 0-547458 0-523766 0.124820 0-491591 0°489268 0-489003 0:481569 0-450449 0-421966 0.391011 0-346339 0.323332 0.300123 O'126590 0-000633 } 700 Note.- 7000 grains make 1 Ib. avoirdapois, and 5760 grains make 1 lb troy; therefore, as 1 lb. avoirdupois : 1 lb. troy :: 7000 : 5760, or as 700 : 576 576, consequently, 1 lb. avoirdupois, multiplied by gives 1 lb. troy, and 700 700 1 lb. troy, multiplied by 576 gives i lb. avoirdupois. For example, 18 ounces is a pound troy; then X12=14:58f the number of troy ounces in one 576 ponnd avoirdupois, and 14:584 multiplied into any number ander Wt. Ib. ar. in the Table, will give its opposite number under Wt. oz. tr.; on the contrary, 567 -be multiplied by any number under Wt. oz. tr., it will produce its 12 X 700 opposite or borizontal number under Wt. Ib. av. Prob. 1. The weight of a body being given, to find its solidity. Divide the given weight in pounds avoirdupois, by the tabular weight corresponding to the name of the same kind, and the quotient will be the solidity in cubic inches ; and if the quotient is divided by 1728, you will have the number of cubic feet. Ex. What is the solidity of a block of marble, weighing 8 tons 14 cwt, in cubic feet? Now 8 tons 14 cwt.=19488 lb. 19448 Theo 0977286 +1728=115'4 cubic feet the solidity. |