--- title: "Skip distance and MUF (curved earth)" subject: "Propagation and Antenna" chapter: 4 key: "semester_6/propagation_and_antenna:skip-muf-curved-earth" source: https://www.bctengineer.com/q/semester_6/propagation_and_antenna/skip-muf-curved-earth site: BCT Engineering --- # Skip distance and MUF (curved earth) Propagation and Antenna · Chapter 4: Propagation and Radio Frequency Spectrum - Asked in: 2079 Chaitra, 7 marks - Video: [Sky Wave Propagation 4 | Relationship Between Skip Distance and MUF for Curved Earth | Antenna Hindi](https://www.youtube.com/watch?v=YeMWoDf3Tn0) — Dopamine - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/propagation_and_antenna/skip-muf-curved-earth - Whole chapter: https://www.bctengineer.com/md/semester_6/propagation_and_antenna/chapter-4 ## Question Derive the expression between skip distance and maximum useable frequency (MUF) assuming earth's curvature. ## Short answer With spherical earth radius R and ionosphere height h, MUF f_MUF = fc sec φ, where tan φ = R sin(D/(2R)) / (R + h - R cos(D/(2R))). For small D, f_MUF ≈ fc √[1 + ((D/2)/(h + D^2/(8R)))^2]. Maximum MUF is fc (R+h)/R, at D_max = 2R arccos(R/(R+h)). ## Answer **Skip distance (D)** is the minimum ground distance at which a sky wave of a given frequency returns to earth; **MUF** is the highest frequency returned for a given distance. Over long paths the earth's curvature must be included, because it limits the angle of incidence at the ionosphere. **Assumptions:** spherical earth of radius R; thin reflecting layer at virtual height h; transmitter T and receiver at ground level; reflection point P above the mid-point; θ = half the angle subtended at the earth's centre. ```text 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: ```text 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: ```text 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: ```text 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: ```text 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): ```text 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. **Example:** fc = 8 MHz, h = 300 km, D = 2000 km (R = 6370 km): θ = 8.99°, tan φ = 2.632, f_MUF = 8 × √(1 + 2.632²) = 22.5 MHz. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In the curved-earth derivation, tan(phi) is found as: A. R sin(theta) / (R + h - R cos(theta)), with theta = D/2R B. R cos(theta) / (R + h) C. h / (D/2) D. R sin(theta) / (R + h) Answer: A. In triangle TQP, TQ = R sin(theta) and QP = (R+h) - R cos(theta). (Key point: Geometry) 2. For D small compared to R, tan(phi) becomes approximately: A. (D/2)/h B. (D/2) / (h + D^2/8R) C. D / (h - D^2/8R) D. (D/2R)/h Answer: B. Using sin(theta) = theta and cos(theta) = 1 - theta^2/2. (Key point: Small-angle approximation) 3. As R tends to infinity, the curved-earth skip distance: A. Becomes zero B. Becomes infinite C. Reduces to D = 2h sqrt[(f/fc)^2 - 1] D. Becomes D = h sqrt[(f/fc)^2 - 1] Answer: C. sqrt(1 - x) is about 1 - x/2, giving D = 2kh. (Key point: Flat-earth check) 4. The curvature limits the single-hop MUF to about: A. 10 fc B. 1.5 fc C. fc exactly D. 3.4 fc, when the ray leaves along the horizon (sin phi_max = R/(R+h)) Answer: D. For h = 300 km, phi_max = 72.8 deg and sec phi_max = 3.37. 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