--- title: "Skip distance derivation (flat earth)" subject: "Propagation and Antenna" chapter: 4 key: "semester_6/propagation_and_antenna:skip-distance-flat-earth" source: https://www.bctengineer.com/q/semester_6/propagation_and_antenna/skip-distance-flat-earth site: BCT Engineering --- # Skip distance derivation (flat earth) Propagation and Antenna · Chapter 4: Propagation and Radio Frequency Spectrum - Asked in: 2075 Baisakh, 8 marks; 2074 Bhadra, 7 marks (asked 2 times) - Video: [Skip Distance (Definition, Basics, Derivation, Formula & Calculation) Explained](https://www.youtube.com/watch?v=JoV6IAyOxEA) — Engineering Funda - 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-distance-flat-earth - Whole chapter: https://www.bctengineer.com/md/semester_6/propagation_and_antenna/chapter-4 ## Question What is skip distance? Derive the expression for skip distance (D) and its relationship with critical frequency (fcr) assuming flat earth surface for both antennas. ## Short answer Skip distance is the minimum ground distance where a sky wave of frequency > fcr returns. For flat earth, D = 2h sqrt((f/fcr)^2 - 1) with h the layer virtual height; if f ≤ fcr, D = 0 (no skip). ## Answer **Skip distance (D)** is the shortest distance, measured along the ground from the transmitter, at which a sky wave of a given frequency (above the critical frequency) comes back to earth after reflection from the ionosphere. Rays sent at steeper angles pass through the layer; the ray at the limiting angle lands at the skip distance. Closer points (beyond the ground-wave range) get no signal: this is the **skip zone**. ```text escapes \ | / ~~~~~~~~~~~~\~~~|~~/~~~~~~~~~~~~~ layer \ /\ \_____ limiting ray / \ \___ ------------T----------------R------ |<-- skip D --->| ``` #### Derivation (flat earth, both antennas on the ground) ```text 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): ```text n = √(1 − 81N / f²) ``` By Snell's law, a wave entering at angle θi turns back where the ray becomes horizontal (refraction angle 90°): ```text 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 ```text f_MUF = fc sec θi (secant law) ``` From the flat-earth geometry (triangle TMP): ```text 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: ```text (f_MUF / fc)² = 1 + D² / 4h² D = 2h √[(f_MUF / fc)² − 1] ``` Here f_MUF is the operating frequency *f* for which the given distance is the skip distance, and fc = fcr. Hence: ```text D = 2h √[(f / fcr)² − 1] ``` #### Relationship with critical frequency - fcr = 9√Nmax depends on maximum electron density of the layer. - For f ≤ fcr: D = 0 (all rays return; no skip zone). - For f > fcr: D grows with the ratio f/fcr. - Higher fcr (daytime, high sunspot activity) gives a shorter skip for the same frequency. **Example:** h = 300 km, fcr = 5 MHz, f = 10 MHz: D = 600 × √3 = 1039 km. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the skip distance? A. The shortest ground distance from the transmitter at which a sky wave of a given frequency returns to earth B. The longest distance a ground wave can travel C. The height of the ionospheric layer above the transmitter D. The distance between two successive hops Answer: A. Rays at steeper angles pass through the layer; the limiting ray lands at the skip distance, and closer points beyond ground-wave range form the skip zone. (Key point: Definition of skip distance) 2. In the derivation, the ray turns back at the point where: A. The refractive index becomes greater than 1 B. The ray hits the earth C. The refraction angle becomes 90 degrees, so sin(theta_i) = n D. The electron density is zero Answer: C. The ray becomes horizontal in the layer, so by Snell's law sin(theta_i) = n = sqrt(1 - 81Nmax/f^2). (Key point: Snell's law condition) 3. For flat earth with virtual height h, tan(theta_i) equals: A. D/h B. (D/2)/h C. h/(D/2) D. 2D/h Answer: B. Triangle TMP has base TM = D/2 and height h, so tan(theta_i) = (D/2)/h. (Key point: Flat-earth geometry) 4. The skip distance formula obtained for flat earth is: A. D = h sqrt[(f/fcr)^2 - 1] B. D = 2h sqrt[1 - (f/fcr)^2] C. D = 2h (f/fcr) D. D = 2h sqrt[(f/fcr)^2 - 1] Answer: D. Using sec(theta_i) = sqrt(1 + D^2/4h^2) = f/fcr and solving for D gives D = 2h sqrt[(f/fcr)^2 - 1]. 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