Critical frequency and skip distance (flat earth)

Critical frequency (fcr) is the highest frequency reflected by a layer at vertical incidence, fcr = 9√Nmax. Skip distance (D) is the minimum ground distance at which a sky wave of frequency f (> fcr) returns to earth. The relation follows from the refraction condition and flat-earth geometry.

Assumptions: flat earth and flat ionosphere; reflection at virtual height h; transmitter and receiver both on the ground (heights neglected); 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]

Here f_MUF is the working frequency f for which D is the skip distance. So the required relation is:

D = 2h √[(f / fcr)² − 1]       or
fcr = f / √(1 + D² / 4h²) = 2hf / √(4h² + D²)

Observations

• If f = fcr, D = 0: the wave returns even straight overhead, so there is no skip zone.
• For f > fcr, D increases with f; a higher frequency needs a more oblique ray to return.
• A higher layer (bigger h) gives a larger skip distance for the same f/fcr ratio.
• At night fcr falls (lower Nmax), so the skip distance for the same frequency increases; operators lower the frequency at night.

Example: h = 250 km, fcr = 6 MHz, f = 12 MHz: D = 2 × 250 × √(2² − 1) = 500 × 1.732 = 866 km.