140

Through C draw CN parallel to HF,meeting FP in N. Then by Art. 8., wehave t y— cz or by substituting c-j-x for t,(c-j-ir) y — cs, or xy = cz- cy , But xy isthe fluxion of the external area, CFN, con-sequently the area CFN = cz-cy ; hencethat of the internal area CFH = CNFH -CFN=ir?/ - {cz - cy) —cy-\-xy - cz =c-\^x[y

— cz — ty - cz — PF.FH - CR-CF, whichwas to be found.

27* Problem. To find the radius of cur-vature to any point in the catenary.

From art. 8., we have y — and % —

~or by making a? = 1 ,y — ~=cz~* and z==

hence- t/=^ and consequently

ji . cx _ t.—JL x x

^3 • z ~ -z 3 ~ CZ 2 3 C t CtZ i C C

= the radius of curvature required.

28. Corollary. At the vertex x==o,and the radius of curvature ^S±fL j n that

C

c * *

case becomes = — = c = the tension atthat point.