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Space wave over plane and actual earth

Short answer

Space wave propagates via direct and ground-reflected waves. On plane earth, field strength decreases as 1/d² and forms lobes; antenna height helps. On actual curved earth, the radio horizon limits range, effective heights are reduced, the reflected wave is weakened by divergence, and a shadow zone appears beyond the horizon.

Explain the effect of space wave propagation on the ground of plane and actual earth.

Space wave propagation is the mode in which the signal reaches the receiver by the direct wave plus the ground-reflected wave through the troposphere. It is used at VHF, UHF and microwaves. The ground affects it in two ways: by reflection (interference of the two waves) and, for real earth, by curvature.

(a) Over a plane (flat) earth

   Tx                          Rx
   |\ ------ direct r1 -------> |
 ht|  \                      / | hr
   |    \  reflected r2    /   |
 ==|======\______P______/======|==
   |<------------ d ---------->|
  • Path lengths: r₁ = √(d² + (ht − hr)²), r₂ = √(d² + (ht + hr)²).
  • For d ≫ ht, hr: path difference Δr = r₂ − r₁ ≈ 2ht·hr/d.
  • Phase difference = (2π/λ)(2ht·hr/d) + π (reflection at grazing angle gives coefficient ≈ −1).
  • Resultant field (E₀/d = field of direct wave at distance d):
E = (2E₀/d) · sin(2π ht hr / λd)
For large d (small angle):  sin x ≈ x
E ≈ 4π E₀ ht hr / (λ d²)
In practical form: E = 88 √P · ht · hr / (λ d²) V/m
  (P in W, heights, λ and d in metres)

Effects:

  • Field strength varies as 1/d² (not 1/d) at large distance, so it falls fast.
  • Near the transmitter, maxima and minima (lobes) occur as d changes, due to interference.
  • Raising either antenna increases the field directly (E ∝ ht·hr), so high masts are used.
  • Horizontal and vertical polarisation behave similarly at grazing incidence over most ground.

(b) Over the actual (curved) earth

  1. Radio horizon: curvature limits the line-of-sight range. With standard refraction (effective radius 4R/3):
d = √(2kR ht) + √(2kR hr) ≈ 4.12 (√ht + √hr) km   (h in m)
  1. Effective antenna heights: the curved ground rises between the antennas, so heights measured from the tangent plane at the reflection point are reduced:
h't = ht − d1²/(2kR),   h'r = hr − d2²/(2kR)

These reduced heights are used in the field formula, giving a weaker field than for plane earth. 3. Divergence factor: reflection from a convex surface spreads the reflected rays, so the reflected wave is weaker; the reflection coefficient is multiplied by a divergence factor D < 1. 4. Shadow zone: beyond the horizon, only weak diffracted fields exist; the field falls rapidly. 5. Tropospheric refraction: decrease of refractive index with height bends rays down and extends the horizon by about 15% (k = 4/3); abnormal conditions give super-refraction, ducting or sub-refraction. 6. Terrain, obstacles and roughness: hills and buildings block or diffract the wave; rough ground gives diffuse rather than mirror reflection (Rayleigh criterion); Fresnel-zone clearance (≥ 0.6 of first zone) is needed for a clean link.

Conclusion of comparison: flat-earth analysis fits short links; for longer links, curvature reduces effective heights, limits range to the radio horizon, and weakens the reflected wave.

✓ Answer checked by Claude Opus 5.5 (high effort)

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