Calculation
of sag and tension
Supports
at the same level
Figure below shows a wire AOB
of length l
supported at two towers A
and B
and are spaced L
unit apart.
Let
O
is the lowest point
of the wire. Consider a length OP
of the curve length s.
If w=weight/unit
length, H=tension
at point O and T=tension
at point P, the tension T can be resolved into horizontal
and vertical
components as :
Then
In
triangle shown in figure below,
ds
represents very small section and therefore we have :
or
and
Substituting
the value of tan(θ),
we get
and
Integrating
both sides, we have
Where
A
is the integration constant. With initial condition at x=0,
s=0
we find A=0.
Therefore
and
At
x=L/2,
s=l/2,
we obtain
Expanding
the sinh(wL/2H)
and ignoring higher order
terms, we get
From
above equations, we can get
Integrating
both sides, we obtain
Where
B
is the integration constant and can be obtained with initial
condition at x=0,
y=o.
Thus, we get B=–H/w.
Therefore
It
is equation
of the sag that is called a catenary.
Source:
Electric
Power Generation, Transmission and Distribution By S.
N. Singh (pages 229,230,231) 
