---
title: "LOS distance: 45 m and 25 m antennas"
subject: "Propagation and Antenna"
chapter: 4
key: "semester_6/propagation_and_antenna:los-distance-45m-25m"
source: https://www.bctengineer.com/q/semester_6/propagation_and_antenna/los-distance-45m-25m
site: BCT Engineering
---

# LOS distance: 45 m and 25 m antennas

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

- Asked in: 2072 Asoj, 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-45m-25m
- Whole chapter: https://www.bctengineer.com/md/semester_6/propagation_and_antenna/chapter-4

## Question

Find out the line of sight distance between the transmitting antenna and receiving antenna if the transmitting antenna height is 45 m and the receiving antenna height is 25 m. [Given: the radius of Earth is 6,378 km].

## Short answer

Using d = √(2R ht) + √(2R hr) with R=6378 km, d = √(2×6.378×10⁶×45) + √(2×6.378×10⁶×25) ≈ 41.8 km.

## Answer

The line-of-sight (LOS) distance is the sum of the distances from each antenna to the horizon. From the geometry of a tangent to a sphere of radius R:

```text
   Tx                         Rx
   |\                         /|
 ht| \         d1   d2       / |hr
   |  `-.______  P  ______.-'  |
         earth (radius R)
 (R + h)² = R² + d²  →  d = √(2Rh + h²) ≈ √(2Rh)
```

**Given:** ht = 45 m, hr = 25 m, R = 6378 km = 6.378 × 10⁶ m.

```text
d1 = √(2 R ht) = √(2 × 6.378×10⁶ × 45)
   = √(5.740×10⁸) = 23 959 m = 23.96 km
d2 = √(2 R hr) = √(2 × 6.378×10⁶ × 25)
   = √(3.189×10⁸) = 17 858 m = 17.86 km
d  = d1 + d2 = 23.96 + 17.86 = 41.82 km
```

**Answer: LOS distance ≈ 41.8 km** (using the given geometric earth radius).

Note: if standard atmospheric refraction is included (effective radius 4/3 × 6378 = 8504 km), d = 27.67 + 20.62 = 48.29 km.

## Check yourself

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

1. The radio horizon distance of one antenna of height h on an earth of radius R is approximately:
   A. h/R
   B. sqrt(2Rh)
   C. 2Rh
   D. sqrt(R/h)
   Answer: B. From (R+h)^2 = R^2 + d^2, d = sqrt(2Rh + h^2) which is approximately sqrt(2Rh). (Key point: Horizon formula)

2. The LOS distance between the two antennas is:
   A. The horizon of the taller antenna only
   B. d1 + d2, the sum of the two horizons
   C. d1 - d2
   D. sqrt(d1 x d2)
   Answer: B. Each antenna sees its horizon; the total is the sum. (Key point: Total LOS)

3. For ht = 45 m, hr = 25 m, R = 6378 km, d is about:
   A. 23.96 km
   B. 17.86 km
   C. 41.8 km
   D. 48.3 km
   Answer: C. d1 = 23.96 km, d2 = 17.86 km, sum = 41.82 km. (Key point: Numerical)

4. If standard refraction is included, how does the distance change?
   A. It decreases
   B. It stays the same
   C. It increases because effective radius is 4/3 R
   D. It becomes zero
   Answer: C. Using effective radius 8504 km gives about 48.3 km. (Key point: Refraction effect)

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