---
title: "Most economical span from sag table"
subject: "Transmission and Distribution System Design"
chapter: 4
key: "semester_8/transmission_and_distribution_system_design:economic-span-sag-table-28mm"
source: https://www.bctengineer.com/q/semester_8/transmission_and_distribution_system_design/economic-span-sag-table-28mm
site: BCT Engineering
---

# Most economical span from sag table

Transmission and Distribution System Design · Chapter 4: Conductor and Support Selection

- Asked in: 2075 Bhadra, 10 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_8/transmission_and_distribution_system_design/economic-span-sag-table-28mm
- Whole chapter: https://www.bctengineer.com/md/semester_8/transmission_and_distribution_system_design/chapter-4

## Question

For the design of most economical span of a high voltage transmission line, following data are available for a particular conductor. Select the most economical span. Given that: conductor diameter = 28 mm, Cross sectional area = 463 mm², UTS = 15910 kg, Wind pressure = 100 kg/m², Factor of safety = 2.

| Span (m) | Maximum sag (m) |
|---|---|
| 250 | 3.24 |
| 275 | 3.88 |
| 300 | 4.57 |
| 325 | 5.32 |
| 350 | 6.12 |

[Figure: conductor arrangement/air clearance on the tower — three conductors with vertical spacing of 2 m between the top and middle and 2 m between the middle and bottom; the top and bottom conductors on one side and the middle conductor on the other, with a horizontal width of 8.7 m across.] Assume: All towers are straight line towers. The air clearance is shown in figure. Minimum ground clearance is 6 m. Neglect the effect of ground wire.

## Short answer

Using wind load on conductors and Ryle's formula for tower weight, tower steel per km is computed for each span. The 250 m span gives the lowest relative tower cost, making it the most economical span.

## Answer

The tower height is built up from the ground clearance, the sag and the conductor spacing. The tower weight is then found from the overturning moment of the wind load with Ryle's formula, and the total tower steel per km is compared.

#### Tower heights
With 6 m minimum ground clearance and conductors 2 m apart vertically (from the figure), and the ground wire neglected:

| Span (m) | Max sag (m) | $h_1 = 6 + S_{max}$ | $h_2 = h_1 + 2$ | $h_3 = H_t = h_1 + 4$ |
|---|---|---|---|---|
| 250 | 3.24 | 9.24 | 11.24 | 13.24 |
| 275 | 3.88 | 9.88 | 11.88 | 13.88 |
| 300 | 4.57 | 10.57 | 12.57 | 14.57 |
| 325 | 5.32 | 11.32 | 13.32 | 15.32 |
| 350 | 6.12 | 12.12 | 14.12 | 16.12 |

Working tension $T = \text{UTS}/\text{FOS} = 15910/2 = 7955$ kg. All towers are straight-line towers, so there is no deviation load. $T$ is needed only to check that the given sags are consistent with it. The 8.7 m horizontal width is the same for every span, so it does not affect the comparison.

**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.

#### Wind load on each conductor ($p = 100$ kg/m², $d = 0.028$ m)
| Span (m) | $F_c = p\,d_c\,S$ (kg) | $F_e = p\,d_e\,S$ (kg) |
|---|---|---|
| 250 | 700.00 | - |
| 275 | 770.00 | - |
| 300 | 840.00 | - |
| 325 | 910.00 | - |
| 350 | 980.00 | - |

#### Overturning moment and tower weight index
$M = F_c(h_1 + h_2 + h_3)$. For example, at 250 m: $M = 700 \times (9.24 + 11.24 + 13.24) = 23{,}604$ kg·m and $H_t\sqrt{M} = 13.24\sqrt{23604} = 2{,}034.1$.

| Span (m) | Tower type (θ) | $M_c$ (kg·m) | $M_e$ (kg·m) | $M$ (kg·m) | $H_t\sqrt{M}$ |
|---|---|---|---|---|---|
| 250 | straight (0°) | 23,604.0 | - | 23,604.0 | 2,034.1 |
| 275 | straight (0°) | 27,442.8 | - | 27,442.8 | 2,299.3 |
| 300 | straight (0°) | 31,676.4 | - | 31,676.4 | 2,593.1 |
| 325 | straight (0°) | 36,363.6 | - | 36,363.6 | 2,921.4 |
| 350 | straight (0°) | 41,512.8 | - | 41,512.8 | 3,284.4 |

#### Tower steel per km
| Span (m) | Weighted $H_t\sqrt{M}$ | Towers per km $= 1000/S$ | Relative cost per km |
|---|---|---|---|
| 250 | 2,034.1 | 4.000 | 8,136.6 |
| 275 | 2,299.3 | 3.636 | 8,361.2 |
| 300 | 2,593.1 | 3.333 | 8,643.8 |
| 325 | 2,921.4 | 3.077 | 8,988.9 |
| 350 | 3,284.4 | 2.857 | 9,384.0 |

Here the sag rises quickly with span (3.24 m to 6.12 m), so both tower height and moment grow faster than the number of towers falls.

**Answer: the most economical span is 250 m** (lowest relative tower cost, 8,136.6 K per km).

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How is the height of the lowest conductor h1 found?
   A. 6 m ground clearance + maximum sag
   B. 6 m - maximum sag
   C. Span/50
   D. Sag + 2 m
   Answer: A. The lowest point of the sagging conductor must be 6 m above ground, so h1 = 6 + Smax. (Key point: Tower height build-up)

2. Which expression gives tower weight in this comparison?
   A. W = K M/Ht
   B. W = K Ht sqrt(M)
   C. W = K Ht M^2
   D. W = K sqrt(Ht) M
   Answer: B. Ryle's formula W = K Ht sqrt(M) relates weight to height and overturning moment. (Key point: Ryle's formula)

3. For the 250 m span, what is the wind force on one conductor?
   A. 100 x 28 x 250 kg
   B. 100 x 0.028 x 250 = 700 kg
   C. 100 x 0.028 x 3.24 kg
   D. 7955 kg
   Answer: B. F = p d S = 100 x 0.028 x 250 = 700 kg. The tension 7955 kg is not used in the moment for straight towers. (Key point: Wind load on conductor)

4. Why is 250 m the most economical span?
   A. It has the smallest sag only
   B. It has the lowest relative tower cost per km (towers per km x tower weight index)
   C. It has the fewest towers per km
   D. It has the largest tower height
   Answer: B. Cost per km is (1000/S) x Ht sqrt(M); this is lowest, 8136.6, at 250 m because height and moment grow faster than tower count falls. (Key point: Economic span criterion)

## Close variants set in other exams

- [Most economical EHV span from sag table](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Aeconomic-span-sag-table-21mm) — Transmission and Distribution System Design, chapter 4, asked 2×

## Related in Transmission and Distribution System Design

- [Most economical span, double circuit line](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Aeconomic-span-double-circuit) — Transmission and Distribution System Design, chapter 4, asked 6×
- [Most economical conductor from cost data](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Aeconomic-conductor-goat-sheep-zebra-deer) — Transmission and Distribution System Design, chapter 4, asked 3×
- [Most economical span, 180 km single circuit](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Aeconomic-span-180km-single-circuit) — Transmission and Distribution System Design, chapter 4
- [Most economical conductor, five options (Rs 7.2)](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Aeconomic-conductor-five-options-rs) — Transmission and Distribution System Design, chapter 4, asked 3×
- [Most economical conductor, five options (Rs 9)](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Aeconomic-conductor-five-options-kwh) — Transmission and Distribution System Design, chapter 4, asked 3×
- [Economical span, 200 km, 10% angle towers](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Aeconomic-span-200km-angle-towers) — Transmission and Distribution System Design, chapter 4, asked 2×