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MUF and skip distance (flat earth)

Short answer

MUF is the highest frequency returned for a given path. For flat earth: MUF = fcr sqrt(1 + D^2/(4h^2)), where D is path length and h virtual height; skip distance D = 2h sqrt((MUF/fcr)^2 - 1).

Define and derive the formula for maximum usable frequency and skip distance assuming earth is perfectly flat.

Definitions

  • Maximum usable frequency (MUF): the highest frequency that is returned to earth by the ionosphere for a given angle of incidence (that is, for a given distance between two points). Frequencies above the MUF pass through the layer.
  • Skip distance (D): the minimum distance from the transmitter at which a sky wave of a given frequency returns to earth.
  • Critical frequency (fc): the MUF for vertical incidence, fc = 9√Nmax.

Derivation (earth assumed perfectly flat)

Assumptions: flat earth and flat ionospheric layer; sharp reflection at virtual height h; transmitter and receiver at ground level; symmetric path.

          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]

Final formulas

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

Remarks

  • MUF increases with distance (larger θi), so long links can use higher frequencies.
  • The flat-earth result is good for distances up to about 500 km; for longer paths the earth's curvature limits θi, and single-hop MUF reaches about 3–4 × fc.
  • In practice the optimum working frequency OWF ≈ 0.85 MUF is used, to allow for ionospheric changes.

Example: fc = 7 MHz, h = 300 km, D = 1000 km: f_MUF = 7 × √(1 + 1000²/(4 × 300²)) = 7 × √3.778 = 13.6 MHz.

✓ Answer checked by Claude Opus 5.5 (high effort)

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