MUF, critical frequency and skip (curved earth)

Maximum Usable Frequency (MUF)

The MUF is the highest frequency that is returned to earth by the ionosphere between two given points (for a given angle of incidence). Higher frequencies pass through the layer. It is related to the critical frequency fcr (highest frequency reflected at vertical incidence) by the secant law: f_MUF = fcr sec φ. In practice the optimum working frequency OWF ≈ 0.85 MUF is used.

Derivation for curved earth

Assumptions: spherical earth of radius R; layer acting as a thin reflector at virtual height h; both antennas on the ground; reflection point P above the mid-point of the path.

First, the secant law. The refractive index of the layer is n = √(1 − 81N/f²). By Snell's law, a ray entering at angle φ returns where n = sin φ:

1 − 81Nmax/f² = sin² φ
81Nmax/f²     = cos² φ
f = 9√Nmax / cos φ = fcr sec φ,   fcr = 9√Nmax

Then the curved-earth geometry:

               P  (reflection point, height h)
              /|\
             / | \      φ = angle of incidence at P
            /φ |  \
   ~~~~~~~~/~~~|~~~\~~~~~~~ ionosphere
       T  /    |    \  R    (on earth surface)
        \.     |     ./
          `.   |θ  .'       θ = D / (2R)
            `. | .'
               O  (earth centre)
   OT = R,  OP = R + h,  arc TR = D

Step 1: Drop a perpendicular from T onto OP. Its foot Q lies at distance R cos θ from O, and TQ = R sin θ. In triangle TQP:

QP    = (R + h) − R cos θ
tan φ = TQ / QP = R sin θ / (R + h − R cos θ)

Step 2: Secant law for the ionosphere (from n = √(1 − 81N/f²) and Snell's law): f_MUF = fc sec φ, with fc = 9√Nmax. So:

f_MUF = fc √(1 + tan² φ)

f_MUF = fc √{ 1 + [ R sin(D/2R) /
              (R + h − R cos(D/2R)) ]² }

Step 3: For D small compared with R, sin θ ≈ θ = D/2R and cos θ ≈ 1 − θ²/2:

tan φ ≈ (D/2) / (h + D²/8R)
f_MUF ≈ fc √{ 1 + [ (D/2) / (h + D²/8R) ]² }

Step 4: Skip distance for a given frequency f. Let k = √[(f/fc)² − 1] = tan φ:

D/2 = k (h + D²/8R)
(k/8R) D² − D/2 + k h = 0
D = (2R/k) [ 1 − √(1 − 2k²h/R) ]   (smaller root)

Check: as R → ∞, √(1 − x) ≈ 1 − x/2, so D → 2kh = 2h√[(f/fc)² − 1], the flat-earth result.

Step 5: Limit due to curvature. The largest angle φ occurs when the ray leaves the transmitter along the horizon (tangent at T):

sin φmax = R / (R + h)
f_MUF(max) = fc sec φmax
D_max (one hop) = 2R cos⁻¹[R / (R + h)]

For h = 300 km, R = 6370 km: sin φmax = 0.955, φmax = 72.8°, sec φmax = 3.37, D_max ≈ 3836 km. So single-hop MUF cannot exceed about 3.4 fc.

Summary

fcr   = 9 √Nmax
f_MUF = fcr sec φ,
tan φ = R sin(D/2R) / (R + h − R cos(D/2R))
D     = (2R/k)[1 − √(1 − 2k²h/R)],  k = √[(f/fcr)² − 1]

Example: fcr = 8 MHz, h = 300 km, f = 16 MHz: k = √3 = 1.732; D = (2 × 6370/1.732)[1 − √(1 − 2 × 3 × 300/6370)] ≈ 1125 km (the flat-earth formula gives 1039 km).