96 
The N.Z. Journal of Science and Technology. 
[Mar. 
per unit length and s the length OP. These forces meet at Q. Produce 
OQ to meet a vertical through P in N ; then the triangle PQN may be 
taken to represent the triangle of forces, from which it will be seen that the 
following relations hold : — 
PN = CS = T sin PQN, QN = T = T cos PQN, PN/QN = ~= tan PQN 
According to the properties of the catenary, the horizontal component 
of the tension at any point along the wire is constant, and is therefore equal 
to the tension T at 0. Let T be represented by a length c of wire whose 
weight is numerically equal to T ; then T = wc. 
B 
Proceeding to the limit we now have the following equation 
dy o os ojs s 
dx T wc c ’ 
In order to determine y in terms of s we have 
PN o;5 
■(!)• 
whence 
5 -j. dy 
PQ \/w 2 S 2 + oo 2 C 2 Vs 2 -f- C 2 ds 
y = 
Jv 
s 2 + c 5 
ds = V s 2 + c 2 + A 
when y = o, 5 = 0 ; then A = — c and y = V s 2 -f- c 2 — c oi y c = Vs' 2 c 2 
If now the axis of x be displaced to a distance c from the apex, so that 
y — c when x = 0 , we have 
y = Vs 2 -f- ° r y 2 ~ s 2 + ° 2 .(2) 
where y is the ordinate referred to the new axis. 
