Economical span, 200 km, 10% angle towers

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.

Data

• Conductor working tension $T_c = \text{UTS}/\text{FOS} = 8000/2 = 4000$ kg, $d_c = 0.020$ m.
• Earth wire $T_e = 4000$ kg, $d_e = 0.016$ m. Wind pressure $p = 80$ kg/m².
• Single circuit with single earth wire: $n_c = 1$, $n_e = 1$.
• 90% straight towers (no deviation load) and 10% angle towers designed for 15°.
• $N = \lceil 200000/S \rceil + 1$.

Deviation load on the angle towers:

$$F_{dc} = F_{de} = 2 \times 4000 \times \sin 7.5° = 1,044.21\ \text{kg}$$

Wind loads

Moments and tower weight index

Sample calculation (250 m, angle tower):

$$\begin{aligned} M_c &= (400 + 1044.21)(10.245 + 13.555 + 16.865) = 58{,}728.8\ \text{kg·m} \\ M_e &= (320 + 1044.21)(28.815) = 39{,}309.7\ \text{kg·m} \\ H_t\sqrt{M} &= 28.815\sqrt{98038.5} = 9{,}022.3 \end{aligned}$$

Total tower steel for 200 km

Weighted index $= 0.9\,(\text{straight}) + 0.1\,(\text{angle})$:

Answer: the most economical span is 350 m (573 towers: about 516 straight and 57 angle towers). The curve is flat beyond 300 m, so 325–350 m spans are nearly equally economical.