224 KANSAS UNIVERSITY QUARTERLY. 
For this take the ‘‘s-curve”’ plotting, already prepared, as 
shown in Fig. 7. 
Lay off one-half the distance D between the given abutments 
along OX and —1, the length of cord given, upward from M to B. 
I 
—J==s,. 
2 
Join B to O and from P, where BO cuts the s-curve drop a 
perpendicular PK upon OX. 
Draw EF parallel to OM at unit’s distance above it. Draw OF 
to D, when DM is the value of c, the parameter to be used in 
plotting the required catenary. With this value for c the curve 
will have the given length between the fixed abutments at given 
distance, D, apart. 
It only remains to lay off c downward from O on another drawing 
and proceed to construct the catenary by the method already ex- 
plained. 
Proof: The required catenary will, we know from previous con- 
siderations, have an equation for s 
(e is rose Se Py 
== ¢ —6 Cus 
2 
It will be of such proportions that if plotted on Fig. 7 OM and 
MB would be coGrdinates of it, when O is the origin. Manifestly 
we can not substitute 
. 
OM-—x,7 /and-NB==s.; 
for x and s, respectively, and solve for c, as we should do in the 
case of an equation of ordinary kind, owing to form of this equa- 
tion, but the constant c must be determined in order to plot the 
catenary. Andit may now be readily seen that the graphical steps 
just indicated do give c correctly. 
By the equation of the simple s-curve 
I OK OK 
PK=—(e --@ ) 
2 
Now PK some constant c=BM 
and OK» the same constant c—=OM 
by similar triangles. 
