Location of maximum sag, supports at different levels

For supports at different levels, the conductor is approximated by a parabola with its lowest point O. The "maximum sag" from each support is measured down to O.

Let $L$ be the span, $h$ the difference in support levels (B higher), $w$ the weight per metre and $T$ the horizontal tension. Let $x_1$ and $x_2$ be the horizontal distances of O from the lower support A and the higher support B.

Expressions for maximum sag

Taking O as the origin, $y = \dfrac{w x^2}{2T}$, so

$$S_1 = \frac{w x_1^2}{2T}\ \text{(from A)}, \qquad S_2 = \frac{w x_2^2}{2T}\ \text{(from B)}$$

Location of maximum sag

The difference of the sags is the difference in support levels:

$$\begin{aligned} h &= S_2 - S_1 = \frac{w}{2T}\left(x_2^2 - x_1^2\right) \\ &= \frac{w}{2T}(x_2 + x_1)(x_2 - x_1) \end{aligned}$$

Since $x_1 + x_2 = L$:

$$x_2 - x_1 = \frac{2Th}{wL}$$

Adding and subtracting with $x_1 + x_2 = L$:

$$x_1 = \frac{L}{2} - \frac{Th}{wL}, \qquad x_2 = \frac{L}{2} + \frac{Th}{wL}$$

The point of maximum sag therefore lies $\dfrac{Th}{wL}$ from mid-span, towards the lower support.

• With equal supports ($h = 0$): $x_1 = x_2 = L/2$.
• If $h > wL^2/2T$, then $x_1 < 0$ and the lowest point falls outside the span (virtual sag). The lower tower then has uplift.
• Under wind and ice, $w$ is replaced by the resultant load per metre.