VHF Link Behind a Ridge: Mast Height Versus Terrain
A VHF link at 40 MHz must connect a battalion command post to a mobile patrol 28 km away. The command post antenna sits on a 10 m mast on flat ground; the patrol antenna is 2 m above ground. Use the radio-horizon approximation d(km) = 3.57(√h1 + √h2), with heights in metres, and ignore atmospheric refraction beyond this formula. A ridge 60 m high lies 12 km from the command post, directly on the path. The patrol reports that with the current mast the link works only up to 16 km, then fails. The signals officer proposes raising the command post mast to 45 m and switching to a 150 MHz handheld set for the patrol. Mast sections are 5 m each; each added section needs a 2-hour rigging team and a guy-wire anchor. The patrol must be on air within 36 hours.
1. Check whether the current mast and the proposed 45 m mast reach 28 km by the horizon formula, and then test the ridge clearance.
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Current mast: d = 3.57(√10 + √2) = 3.57(3.162 + 1.414) = 3.57 × 4.576 ≈ 16.3 km. This matches the patrol's report of failure beyond 16 km.
45 m mast: d = 3.57(√45 + √2) = 3.57(6.708 + 1.414) = 3.57 × 8.122 ≈ 29.0 km. On the horizon formula alone, 29 km nominally covers the 28 km path.
Terrain check: the straight line from 45 m to 2 m over 28 km falls to 45 − (43 × 12/28) = 45 − 18.4 = 26.6 m above ground at the ridge, which is 12 km out. The ridge is 60 m high, so the path is blocked by about 33 m. The mast therefore does not solve the link; the horizon calculation is necessary but not sufficient.
2. Evaluate the proposal to switch the patrol to a 150 MHz handheld set.
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Raising frequency from 40 MHz to 150 MHz shortens the wavelength and reduces diffraction around an obstacle. A 60 m ridge that already blocks the path will block a higher frequency at least as severely. Free-space loss also rises with frequency, and a handheld has lower power and a poorer antenna than a vehicle set.
The 150 MHz proposal therefore does not fix the 33 m obstruction and probably worsens the link margin. It should be rejected as a solution to the terrain problem, though it may be acceptable later for short-range communication once a relay exists.
3. Recommend a solution that can be operational within 36 hours, with numbers.
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Use a relay on the ridge, which has line of sight to both ends. From the ridge to the command post is 12 km; with a 10 m relay mast and the 10 m command mast, the horizon is 3.57(3.162 + 3.162) ≈ 22.6 km, comfortably more than 12 km. From the ridge to the patrol is 16 km; with a 10 m relay mast and the 2 m patrol antenna, the horizon is 3.57(3.162 + 1.414) ≈ 16.3 km, just sufficient but marginal, so a 15 m relay mast gives 3.57(3.873 + 1.414) ≈ 18.9 km and a proper margin.
Time: a 15 m mast needs three 5 m sections, about 6 hours of rigging, plus transport and radio setup. That fits inside 36 hours. Raising the command mast to 45 m would take 7 sections × 2 h = 14 hours and still fail, so it is wasted effort. Recommend the ridge relay with a 15 m mast, a second radio, and a check that the relay's frequency and power are cleared by higher signals authority.