BY MEANS OF LONG STEEL RIBANDS.. 37 
The third term on the right-hand side being nine times as great 
as the gonad term in the e sponses of the equation to the 
catenary curve. The term is so small, however, that it may be 
neglected. The appreciable pa hencfor re, agree exactly, whether 
the solution be catenary or cir 
Zan? im) 2 
IV.— vaca T-tr- oe 
The following is a strictly accurate solution of the catenary 
curve "a any a angle :— 
-s=L;«#-x=Ilsin{g; y'-y=lecosl. 
c - — ; er = om 4 
I=5 ie 9) so =S (ea. °) 
= <) 3 f= (- —e ) 
—z Ho —z 
Therefore, 7 cos f=5 (z te te °) 
—€°--€ 
A) 
—2 2 -—2£ 
and L=$ (i _ eager <) 
a 
Therefore, by addition Z +1 cos =c (c = » 
and by subtraction LZ —/ cos =c C. ee « 2) 
—a’+ 
by multiplication L’ — Pecos’ f=c* ts — . 3) 
+ (2-2 —(x'—2 
or ¢ f. (S 3) =*) . at 
Isin€ --2sin ¢ 
That is LZ? - P cos’ f=? E = ne 5 ae 
Expanding Z*—P cos’ £=P sin® eo oa + Sapa fe 
? sin HM sin * sin’ ¢ ee re gg & 
12c? 360c* * 50160 
So far the solution is strictly accurate ; but asc can only be 
Therefore, L* = 7 + + &e. 
Jf + be 
nearly parallel to the chord 7. Therefore, r sin {= very approxi- 
mately ; and tsin¢_« Hence the above equation becomes 
_» , fw’ sin f , Mot sin °f r= { S = ts 
we as sep But L?= < 1+(r ae 
*+ 20 (7-1) hs (e-ap Rejecting the final term of this last 5 = 
