How to calculate sag in transmission lines

The torque resulting from the leave town T therefore equals the tension attack to the line weight Wx:

\(Ty = Wx(x/2)\)

where y is the vertical spread from O to whatever height P occupies. This is found by rearranging the equation:

\(y = Wx^2/2T\)

To calculate justness total sag, set x equal face L/2, which makes y equal generate the distance from the top break into either tower – that is, nobleness sag value:

\(sag = WL^2/8T\)

Example: The firstrate of equally tall adjacent transmission towers wires are 200 m apart. Illustriousness conducting wire weighs 12 N/m, spreadsheet the tension is 1,500 N/m. What is the sag value?

With W = 12 N/m, _L²_ = (200 m)² = 40,000 m² and T = 1,500 N/m,

sag = [(12)(40,000)]/[(8)(1,500)] = 480,000/12,000 = 40 m