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A, B, be three points in the proposedcurve.

For CE=c, the angle CEM=<p, and EM~t, corresponding to the common cate-nary. Hence NP=EL=EM:=A and bythe property of the logarithmic curve, CD

(=EN) = c hyp. log.-^-=the absciss of the

proposed curve, and from the nature of theconstruction DA or DB=CR=the lengthof a circular arc, whose radius EC=c, andsecant EM=if; hence DA is the ordinateof the proposed curve.

Draw BT parallel to EM, BT is the tan-gent at the point B. Draw the commoncatenary GCI and through L the ordinateGL. Then GL=BC, and EM = EL =the tension in both catenaries.

The angle LG£=the angle DBT=R,EC.The weight of the chains CB and CG areequal. An ordinate of the catenary BCAcan never exceed one-fourth the circum-ference of a circle, of which EC is theradius.

Let the curve be supposed to be of thethickness Ec at the vertex C, then the