Minimum clearance, towers at 45 m and 70 m

For supports at unequal heights, first find where the lowest point of the conductor lies. If it falls outside the span, the lowest conductor point is at the lower tower.

Data: $w = 0.844$ kg/m, $L = 300$ m, $h = 70-45 = 25$ m. The tension is not given. Assume the conductor is the same as in the companion question (UTS = 8000 kg, factor of safety 2), so $T = 4000$ kg.

Position of the lowest point

Measured from the lower tower TR1:

$$x_1 = \frac{L}{2} - \frac{Th}{wL} = 150 - \frac{4000\times25}{0.844\times300} = 150 - 394.94 = -244.94\ \text{m}$$

$x_1$ is negative, so the vertex lies 244.9 m outside the span, beyond TR1. Inside the span, the conductor rises continuously from TR1 (45 m) to TR2 (70 m).

                               TR2 o 70 m
                          __..--`  |
              TR1 __..--``         |
     vertex  45 m o                |
      (outside)   |<---- 300 m --->|
 ===== reference ===================

Generality of the result

The vertex lies inside the span only if

$$T < \frac{wL^2}{2h} = \frac{0.844\times300^2}{2\times25} = 1519.2\ \text{kg}$$

Any practical working tension is above 1519 kg, since it is usually 20–50% of UTS. So the result does not depend on the exact tension assumed.

Answer: the minimum clearance of the conductor from the reference is 45 m, at tower TR1. The conductor has no dip within the span, because its vertex falls about 245 m outside it (for T = 4000 kg).