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Skip distance derivation (flat earth)

Short answer

Skip distance is the minimum ground distance where a sky wave of frequency > fcr returns. For flat earth, D = 2h sqrt((f/fcr)^2 - 1) with h the layer virtual height; if f ≤ fcr, D = 0 (no skip).

What is skip distance? Derive the expression for skip distance (D) and its relationship with critical frequency (fcr) assuming flat earth surface for both antennas.

Skip distance (D) is the shortest distance, measured along the ground from the transmitter, at which a sky wave of a given frequency (above the critical frequency) comes back to earth after reflection from the ionosphere. Rays sent at steeper angles pass through the layer; the ray at the limiting angle lands at the skip distance. Closer points (beyond the ground-wave range) get no signal: this is the skip zone.

          escapes
             \    |   /
  ~~~~~~~~~~~~\~~~|~~/~~~~~~~~~~~~~ layer
               \ /\  \_____
   limiting ray /  \       \___
  ------------T----------------R------
              |<-- skip D --->|

Derivation (flat earth, both antennas on the ground)

          ionosphere (virtual height h)
  ~~~~~~~~~~~~~~~~~~ P ~~~~~~~~~~~~~~~~~~
                   / | \
                  /θi|  \
                 /   |h  \
                /    |    \
   flat earth  T-----M-----R
               <--- D ---->
  TM = MR = D/2,  θi = angle of incidence

Refractive index of an ionised layer (electron density N per m³, frequency f in Hz):

n = √(1 − 81N / f²)

By Snell's law, a wave entering at angle θi turns back where the ray becomes horizontal (refraction angle 90°):

sin θi = n = √(1 − 81Nmax / f²)
cos² θi = 81Nmax / f²
f = 9√Nmax / cos θi

For vertical incidence (θi = 0) this gives the critical frequency fc = 9√Nmax, so

f_MUF = fc sec θi          (secant law)

From the flat-earth geometry (triangle TMP):

tan θi = (D/2) / h
sec θi = √(1 + tan² θi) = √(1 + D² / 4h²)

f_MUF = fc √(1 + D² / 4h²) = fc √(4h² + D²) / (2h)

Solving for the skip distance:

(f_MUF / fc)² = 1 + D² / 4h²
D = 2h √[(f_MUF / fc)² − 1]

Here f_MUF is the operating frequency f for which the given distance is the skip distance, and fc = fcr. Hence:

D = 2h √[(f / fcr)² − 1]

Relationship with critical frequency

  • fcr = 9√Nmax depends on maximum electron density of the layer.
  • For f ≤ fcr: D = 0 (all rays return; no skip zone).
  • For f > fcr: D grows with the ratio f/fcr.
  • Higher fcr (daytime, high sunspot activity) gives a shorter skip for the same frequency.

Example: h = 300 km, fcr = 5 MHz, f = 10 MHz: D = 600 × √3 = 1039 km.

✓ Answer checked by Claude Opus 5.5 (high effort)

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