Tower cost factors and bending moment at angle

Tower cost is roughly proportional to tower weight. By Ryle's formula, $W_t = K H_t\sqrt{M\times FS}$, so the weight depends on the tower height $H_t$ and on the bending (overturning) moment $M$ at its base.

Factors affecting tower cost

1. Voltage level: a higher voltage needs longer insulator strings and air clearances. This makes the cross arms longer, the conductor spacing larger, the ground clearance higher, and so the tower taller.
2. Span length: a longer span gives a larger sag and a taller tower. It also gives a larger wind span, so the bending moment rises. On the other hand, fewer towers are needed per km.
3. Conductor size and tension: a larger diameter means more wind load. A heavier conductor means a larger vertical load. Higher tension means larger loads at angle points and under broken-wire conditions.
4. Number of circuits and earth wires: a double circuit adds cross-arm levels and doubles the loads on the tower.
5. Line deviation angle: angle towers (type B, C, D) must resist $2T\sin(\alpha/2)$ from every wire, so they are heavier.
6. Climate: wind pressure, ice loading and temperature range.
7. Terrain and soil: foundation type, hill slopes, river crossings needing extra-tall towers, and transport access.
8. Factor of safety and broken-wire conditions in the design code.

Bending moment due to a line deviation of α

The line turns through an angle α at the tower. Each wire pulls with tension $T$ along the two line directions.

  T (to span 1)                          T (to span 2)
       `-._                          _.-`
           `-._         F        _.-`
               `-._     ^    _.-`
                   `-._ | _.-`
                        O   tower
  each pull makes alpha/2 with the original line direction
  and (90 deg - alpha/2) with the bisector OF

1. Without the turn the two pulls are exactly opposite. After turning by α, each pull makes an angle of $(90^\circ-\alpha/2)$ with the bisector of the angle between the two spans.
2. Resolve each pull. The components perpendicular to the bisector are $T\cos(\alpha/2)$ in opposite directions, so they cancel.
3. The components along the bisector are each $T\cos(90^\circ-\alpha/2) = T\sin(\alpha/2)$, in the same direction, so they add:

$$F = T\sin\frac{\alpha}{2} + T\sin\frac{\alpha}{2} = 2T\sin\frac{\alpha}{2}$$

This resultant acts along the bisector, towards the inside of the angle. 4. This force acts at the height $h_i$ of each wire. Wind on each wire, over its wind span $L$, adds $p\,d\,L$. So the bending moment at ground level is:

$$M = \sum_{i}\left(2T_i\sin\frac{\alpha}{2} + p\,d_i\,L\right)h_i$$

The sum covers every power conductor and the earth wire(s). For a double circuit there are two conductors at each of the three levels, and the earth wire acts at the top height $H_t$.

For a straight-line tower, α ≈ 0, so only the wind term remains. For angle towers the deviation term usually dominates. For example, with $T=6000$ kg and $\alpha = 30^\circ$, the deviation force is $2\times6000\sin15^\circ = 3106$ kg per conductor. That is much larger than the wind load on a 300 m span of a 4 cm conductor, which is $100\times0.04\times300 = 1200$ kg.