Problem.-To find the circumference of an ellipse, or an oval hoop or ring. Rule.-Add the length of the two axes together, and multiply the sum by 1.5708 for the circumference; or as it may be used in the Table of Circumferences, take half the sum of the axes as a diameter, with the breadth of the iron added, and enter the Table of Circumferences where it will be found. Ex. Required the circumference of an elliptical · hoop, whose axes are 18 and 13 inches, the thickness of the iron being 2 inches. Entering into the Table of Diameter with 181 inches, the circumference will be found to be 4 feet 9 inches. In constructing elliptical hoops of angled iron, with the angle outside, reference must be made to the Tables for hoops of angled iron: the operation will be similar to the above example. But in hoops where the angle is inside, the thickness of the iron must be taken from half the sum of the axes. Note. It must be observed, that in the examples given in the Observations on Table I., and also on hoops formed of angled iron, that those circumferences are nothing more than the ends of the iron meeting together: therefore every smith must allow for the thickening of the ends of the metal previous to scarving the same in order to weld it. J. F. WEIGHT OF A SUPERFICIAL FOOT OF PLATE OR SHEET IRON, COPPER, AND BRASS, IN POUNDS. Note.-The thickness of the nut is equal the bolt's diameter. 14 3 Note.-No. 1 wire gauge equal ths of an inch. The great variety of thicknesses into which copper is manufactured, causes in trade the weight to be named whereby to determine the thickness required, the unit being that of a common sheet, so designated, viz., 4 feet by 2 feet, in lbs.; thus, A 70lb plate is ths of an inch in thickness. The thickness of lead is also in common determined or understood by the weight, the unit being that of a square or superficial foot; thus, 4-lb lead is th of an inch in thickness. 1. Suppose I have an article of plate iron, the weight of which is 728 lbs., but want the same of copper, and of similar dimensions, what will be its weight? 728 x 1·16-844-48 lbs. 2. A model of dry pine weighing 32 lbs., and in which the iron for its construction forms no material portion of the weight, what may I anticipate its weight to be in cast iron? 32.5 x 11=357.5 lbs. Note.-It frequently occurs in the formation or construction of models, that neither the quality nor condition of the timber can be properly estimated, and in such cases it may be a near enough approximation to reckon 10 lbs. of cast iron to each lb. of model. WEIGHTS OF 9-FEET LENGTHS OF CAST-IRON PIPES, OF VARIOUS DIAMETERS. TO ASCERTAIN THE WEIGHTS OF PIPES OF VARIOUS METALS, AND ANY DIAMETER REQUIRED. Rule. To the interior diameter of the pipe, in inches, add the thickness of the metal; multiply the sum by the decimal numbers opposite the required thickness and under the metal's name; also by the length of the pipe in feet, and the product is the weight of the pipe in lbs. 1. Required the weight of a copper pipe whose interior diameter is 7 inches, its length 6 feet, and the metal of an inch in thickness. 7.5+125=7·625 × 1'52 × 6.25=72.4 lbs. 2. What is the weight of a leaden pipe 181 feet in length, 3 inches interior diameter, and the metal of an inch in thickness? 3+25=3·25 × 3·867 × 18·5=282.5 lbs. |