---
title: "Tropospheric range: 100 ft and 50 ft antennas"
subject: "Propagation and Antenna"
chapter: 4
key: "semester_6/propagation_and_antenna:tropospheric-range-numerical"
source: https://www.bctengineer.com/q/semester_6/propagation_and_antenna/tropospheric-range-numerical
site: BCT Engineering
---

# Tropospheric range: 100 ft and 50 ft antennas

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

- Asked in: 2073 Magh, 5 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/tropospheric-range-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_6/propagation_and_antenna/chapter-4

## Question

Find the maximum range of tropospheric transmission for which the transmitting antenna height is 100 ft and receiving antenna is 50 ft.

## Short answer

Using d (miles) = √(2ht) + √(2hr) with ht = 100 ft, hr = 50 ft: √200 ≈ 14.14, √100 = 10, total 24.14 miles (38.9 km).

## Answer

The maximum range of tropospheric (space-wave) transmission is the sum of the radio horizons of the two antennas. With standard atmospheric refraction (effective earth radius 4/3 R), the horizon in **miles** for a height in **feet** is:

```text
d (miles) = √(2 ht) + √(2 hr)     (h in ft)
```

**Given:** ht = 100 ft, hr = 50 ft.

```text
√(2 × 100) = √200 = 14.142 miles
√(2 × 50)  = √100 = 10.000 miles
d = 14.142 + 10.000 = 24.142 miles
d = 24.142 × 1.609 = 38.85 km
```

Check in metric units: ht = 30.48 m, hr = 15.24 m,
d = 4.12(√30.48 + √15.24) = 4.12(5.521 + 3.904) = 38.83 km (same, within rounding).

**Answer: maximum range ≈ 24.1 miles ≈ 38.9 km.**

## Check yourself

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

1. The maximum range of tropospheric transmission is found as:
   A. The horizon of the taller antenna only
   B. The sum of the radio horizons of the two antennas
   C. The difference of the two horizons
   D. The product of the two horizons
   Answer: B. Maximum range is the transmitter horizon plus the receiver horizon. (Key point: Range concept)

2. With standard refraction (4/3 earth), which formula gives d in miles for heights in feet?
   A. d = sqrt(ht) + sqrt(hr)
   B. d = 4.12(sqrt(ht) + sqrt(hr))
   C. d = sqrt(2ht) + sqrt(2hr)
   D. d = sqrt(2ht x hr)
   Answer: C. For feet and miles the horizon is sqrt(2h), summed for both antennas. (Key point: Formula in feet)

3. For ht = 100 ft and hr = 50 ft, what is the range in miles?
   A. 24.1 miles
   B. 14.1 miles
   C. 34.1 miles
   D. 10.0 miles
   Answer: A. sqrt(200) + sqrt(100) = 14.142 + 10 = 24.14 miles, about 38.9 km. (Key point: Numerical result)

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