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Critical frequency; MUF and skip distance (flat earth)

Short answer

Critical frequency fcr = 9√Nmax is the maximum vertical-incidence frequency reflected by a layer. For flat earth, with virtual height h, MUF = fcr √(1 + D²/4h²) and skip distance D = 2h √[(fMUF/fcr)² − 1].

Define the critical frequency (fcr). Derive an expression for the relation between skip distance and maximum usable frequency (fmuf) in SW propagation assuming that the Earth is flat.

Critical frequency

The critical frequency (fcr) of an ionospheric layer is the highest frequency that is reflected back to earth when the wave is sent vertically upward. Higher frequencies at vertical incidence pass through the layer. It depends on the maximum electron density of the layer:

fcr = 9 √Nmax    (fcr in Hz, Nmax in electrons/m³)

Typical values: E ≈ 3–4 MHz, F2 ≈ 5–12 MHz.

Relation between skip distance and MUF (flat earth)

Skip distance (D) is the shortest distance from the transmitter at which a sky wave of a given frequency returns to earth. The maximum usable frequency (f_MUF) is the highest frequency that returns to earth for a given distance (angle of incidence).

Assumptions: earth and ionosphere are flat; reflection takes place at virtual height h; the path is symmetric (both antennas on 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]

Result

f_MUF = fcr √(1 + D²/4h²)
D     = 2h √[(f_MUF/fcr)² − 1]
  • For f ≤ fcr, D = 0 (no skip zone).
  • As frequency rises above fcr, the skip distance grows.

Example: h = 300 km, fcr = 8 MHz, f = 16 MHz: D = 2 × 300 × √(4 − 1) = 1039 km.

✓ Answer checked by Claude Opus 5.5 (high effort)

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