calculate for volume as though it was solid. = D2x.7854 area of top (or bottom). Area x depth=cubical contents. 20x20x.7854-314.16 sq. inches. 314.16×6=1884.96 cubic inches. FIG. 6. Now deducting the various holes. Central hole of 3 inches diameter by 6 inches long. 3x3x.7854-7.0686 and 7.0686 x 6 = 42.4116 or 42.41 + cubic inches. Taking holes next, we find that a the turret is 20 inches in diameter and central hole 3 inches, the amount of solid metal in line with the diameter must be 20 -3=17 inches; or each hole must be 8% inches long as 8%+3+8%=20 inches, the diameter of turret. A hole 2 inches in diameter by 8% inches long gives 2x2x.7854=3.1416x 8.5 26.70 cubic inches each. Six holes will give 6×26.70=160.20 cubic inches for the six holes. Collecting the different volumes to be deducted, we have 42.41 cubic inches for central hole, plus 160.20 cubic inches for the six tool holes gives 42.41+160.20-202.61 cubic inches to be deducted from 1884.96 cubic inches. 1884.96-202.61=1682.35 cubic inches, the total volume of turret. Taking cast iron at .26 pounds per cubic inch we have 1682.35×.26=437.51 pounds. This is not absolutely correct owing to the slight difference between the curvature of the outer diameter of turret and the central hole, but the difference is so small as to be of little consequence. The ball or sphere is not often met with in practice, but it is well to know how to calculate its surface and contents. The surface of a sphere is found by multiplying the area of a circle of the same diameter by 4. A 10 inch sphere or ball will have a surface equal to four times the area of a circle of the same diameter, or in formula language: D2x.7854×4 or (D2×3 1416) =surface of sphere. 10x10x.7854= = 78.54x4 314.16 square inches, as surface of a 10 inch ball. The cubical contents or volume of a sphere is found by multiplying the surface by the radius or by the diameter. In our 10 inch sphere we have 314.16 square inches multiplied by of 10 or 13 =523.6 cubic inches. Another and perhaps easier rule is to multiply the cube of the diameter by "pi" (or .5236, as 3.1416÷6=.5236). This gives diameter cubed or 10×10 × 10=1,000 × 5236=523 6 cubic inches as before. The cone is another form which it is sometimes handy to know about, as for example the lathe center you use in everyday work. You can easily calculate the surface and weight of all except the point, but how shall we find that except by guessing at it? The wise men of old solved that as they did much else that we use without thanking them. The curved (convex in this case) surface of a cone equals the circumference of the base multiplied by half the slant height or side. Take a lathe center one inch in diameter at the base and if it is ground to 60 degrees the slant side will also be one inch. Then 1x3.1416-3.1416x2(% of 1) 1.5708 square inches surface. The volume or cubical contents equals the area of the base multiplied by the vertical height, not the slant height. Here we must use our triangle formula to find the height, and we have the hypothenuse as 1 inch, the base as 1⁄2 inch, so we say 1x 1—.5x.5=1—.25.75. Square root of .75.866. Then the area of base equals 1 x 1 x .7854.7854; .7854 × 3 of 866.7854.288=.2261 of a cubic inch. Take a block of steel like figure 7, and estimate its weight, the dimensions. being as given in the figure. This is simply a case of careful calculation and it can best be done dividing it into sections as shown by the dotted lines and computing each section, separately, noting and adding each section and not forgetting to deduct the half-round groove groove), A one inch circle is 1 x 1 x.7854-.7854 square inches x 8=6.2832 cubic inches, and as the groove is half round we divide this by 2 making 3.1416 to deduct from 252 248 8584 cubic inches. = A cubic inch of steel weighs .283 pounds and 248.85 x .283-70.424 pounds. The following table of metals giving the weights per cubic inch may be useful in calculations of this kind. |