---
title: "LOS distance: 125 m and 9 m antennas"
subject: "Propagation and Antenna"
chapter: 4
key: "semester_6/propagation_and_antenna:los-distance-tv-transmitter"
source: https://www.bctengineer.com/q/semester_6/propagation_and_antenna/los-distance-tv-transmitter
site: BCT Engineering
---

# LOS distance: 125 m and 9 m antennas

Propagation and Antenna · Chapter 4: Propagation and Radio Frequency Spectrum

- Asked in: 2074 Bhadra, 6 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/propagation_and_antenna/los-distance-tv-transmitter
- Whole chapter: https://www.bctengineer.com/md/semester_6/propagation_and_antenna/chapter-4

## Question

The antenna of a TV transmitter is located at a height of 125 m above ground level. Calculate the distance up to which the LOS communication is possible if the height of receiving antenna is to be 9 m.

## Short answer

Using d = 4.12(√ht+√hr) km with 4/3 earth radius, ht = 125 m, hr = 9 m: √125 = 11.18, √9 = 3, so d = 4.12 × 14.18 ≈ 58.4 km.

## Answer

For space-wave (line-of-sight) communication, the maximum distance is the sum of the radio horizons of the two antennas. With standard atmospheric refraction, the effective earth radius is kR with k = 4/3.

**Given:** ht = 125 m, hr = 9 m, R = 6370 km, k = 4/3.

**Formula:**
```text
d = √(2kR·ht) + √(2kR·hr)
With kR = 8493 km and h in metres:
d (km) = 4.12 (√ht + √hr)
```

**Calculation:**
```text
√ht = √125 = 11.180
√hr = √9   = 3.000
d   = 4.12 × (11.180 + 3.000)
    = 4.12 × 14.180
    = 58.42 km
```
Separately: transmitter horizon = 4.12 × 11.180 = 46.06 km, receiver horizon = 4.12 × 3 = 12.36 km.

(If refraction is ignored, d = 3.57(√ht + √hr) = 50.62 km.)

**Answer: LOS communication is possible up to about 58.4 km** (taking standard 4/3 earth refraction).

## Check yourself

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

1. The LOS distance with standard refraction (k = 4/3) is:
   A. d = 4.12 (sqrt(ht) + sqrt(hr)) km, h in m
   B. d = 3.57 (sqrt(ht) x sqrt(hr)) km
   C. d = 4.12 (ht + hr) km
   D. d = 4.12 sqrt(ht + hr) km
   Answer: A. It is the sum of the two radio horizons. (Key point: LOS formula)

2. The constant 4.12 (instead of 3.57) is used because:
   A. The earth is flat
   B. Tropospheric refraction makes the effective earth radius 4R/3
   C. Antennas are higher
   D. The frequency is higher
   Answer: B. It extends the horizon by about 15%. (Key point: Why 4.12)

3. For ht = 125 m and hr = 9 m, the LOS distance is about:
   A. 50.6 km
   B. 46.1 km
   C. 58.4 km
   D. 72.6 km
   Answer: C. 4.12 x (11.18 + 3) = 58.4 km. (Key point: Calculation)

4. The transmitter's own radio horizon (ht = 125 m) is about:
   A. 46 km
   B. 12 km
   C. 58 km
   D. 11 km
   Answer: A. 4.12 x 11.18 = 46.06 km. (Key point: Transmitter horizon)

## Related in Propagation and Antenna

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