100

numbers .01543 and .17542, and multiplyeach of them by the tension at the vertex183.8 and the results will be 2.836 and32.24. Lay off AD=.2.836, and fromD parallel to B C lay off DE=32.24,then shall E be a point in the -curve. As-sume <p again equal to any other numberof degrees (which is contained in the table)as 18°„40', and from columns os and y oppo-site <p lS^AO' take out the numbers .05555and .33171, which multiply by the tensionat the vertex, and the results will be 10.21and 60.97. Lay off AF=10.21, andthrough F parallel to BC lay off FG=60.97, then shall G be another point in thecurve. In the same manner, by assumingdifferent values of <p in succession we mayfind as many points I, L, &c. in the curveas we please ; and hence a curve AEGILC, drawn through all the points so found,will be the catenary required.