Most economical span, 180 km single circuit

Method. For each span, the transverse load at the tower comes from two sources:

• wind on the conductors and earth wires: $F = p \times d \times S$, taking the wind span equal to the span $S$ and using the full projected area;
• the resultant of the conductor tensions at a line deviation $\theta$: $F_d = 2T\sin(\theta/2)$.

These loads act at the conductor heights. Their overturning moment at ground level is

$$M = n_c\,(F_c + F_{dc})(h_1 + h_2 + h_3) + n_e\,(F_e + F_{de})\,H_t$$

where $n_c$ is the number of conductors at each level and $n_e$ the number of earth wires. The tower weight is found from Ryle's formula

$$W = K\,H_t\sqrt{M}$$

$K$ is the same for all spans, so it cancels in the comparison. Tower cost is proportional to tower weight, so the span that gives the smallest total tower weight is the most economical span (with the conductor already fixed). Wind on the tower body is neglected.

Assumptions

• Single circuit: one conductor at each of $H_1$, $H_2$, $H_3$ ($n_c = 1$), and one earth wire at $H_t$ ($n_e = 1$).
• Standard tower types (IS 802 practice): type A is a suspension tower for 0°–2° deviation and type B is a small-angle tower for 2°–15°. Each type is designed for its maximum angle.
• Number of towers for 180 km: $N = \lceil 180000/S \rceil + 1$, with 80% type A and 20% type B.

Deviation loads

Wind loads ($p = 100$ kg/m²)

Moments and tower weight index

Sample calculation (250 m, type A):

$$\begin{aligned} M_c &= (630 + 239.45)(10.22 + 13.60 + 16.98) = 35{,}473.4\ \text{kg·m} \\ M_e &= (410 + 99.13)(28.84) = 14{,}683.3\ \text{kg·m} \\ H_t\sqrt{M} &= 28.84\sqrt{50156.7} = 6{,}458.9 \end{aligned}$$

Total tower steel for the line

Weighted index $= 0.8(H_t\sqrt{M})_A + 0.2(H_t\sqrt{M})_B$, multiplied by the number of towers $N$:

The total tower weight keeps falling up to the largest span tabulated.

Answer: the most economical span is 350 m (lowest total tower weight index, 4.846 × 10⁶ K, with 516 towers).