12.1 Line-of-Sight, Tropospheric Ducting, and Diffraction

12.1 Line-of-Sight, Tropospheric Ducting, and Diffraction

At high frequencies (HF, $3\text{ to }30\text{ MHz}$), radio waves rely extensively on ionospheric refraction to travel thousands of kilometres beyond the curvature of the Earth. However, as operating frequencies transition into the Very High Frequency (VHF, $30\text{ to }300\text{ MHz}$), Ultra High Frequency (UHF, $300\text{ to }3000\text{ MHz}$), and Super High Frequency (SHF / microwave, $3\text{ to }30\text{ GHz}$) regimes, electromagnetic waves routinely penetrate the normal ionospheric layers ($D$, $E$, and $F$) without sufficient refraction to return to ground. Consequently, terrestrial communication at VHF and UHF depends predominantly on propagation through the lowest layer of Earth's atmosphere: the troposphere.

While VHF and UHF transmissions are broadly characterized as "line-of-sight" (LOS) modes, the physical trajectory of a radio wave in the atmosphere is rarely a straight geometric line. Variations in atmospheric temperature, barometric pressure, and water vapour pressure bend, guide, scatter, and diffract electromagnetic energy, establishing communication paths that extend well beyond the visual horizon.

---

Atmospheric Refraction and the Radio Horizon

The troposphere extends from the Earth's surface to an altitude of roughly $10\text{ to }12\text{ km}$ at temperate latitudes. The propagation velocity ($v$) of an electromagnetic wave through air is governed by the atmospheric refractive index ($n$):

$$v = \frac{c}{n}$$

where $c \approx 2.9979 \times 10^8\text{ m/s}$ is the speed of light in vacuum. In air, $n$ exceeds unity by only a tiny fraction (typically $n \approx 1.0003$ at sea level). To handle these microscopic differences conveniently, atmospheric scientists define radio refractivity ($N$) in dimensionless "$N$-units":

$$N = (n - 1) \times 10^6$$

Empirically, atmospheric refractivity depends on three thermodynamic variables:

$$N = \frac{77.6}{T} \left( P + \frac{4810 e}{T} \right)$$

where:

• $P$ is total atmospheric barometric pressure in hectopascals (hPa or mbar),
• $T$ is absolute temperature in Kelvin (K),
• $e$ is the partial pressure of water vapour in hectopascals (hPa).

                     Wavefront Bending in the Troposphere

                Higher Altitude: Lower n -> Faster Wavefront (v_top)
                     ---------------------->
                    /                      /
                   /                      /
                  /                      /   Wavefront tilts downward
                 /                      /    toward Earth's surface
                ----------------------->
                Lower Altitude: Higher n -> Slower Wavefront (v_bottom)

              =======================================================
                                 Earth's Surface

The Standard Refractivity Gradient and the 4/3 Earth Radius

In a well-mixed, "standard" atmosphere, atmospheric pressure and moisture content decrease exponentially with altitude, while temperature drops at an average environmental lapse rate of approximately $6.5^\circ\text{C}$ per kilometre ($6.5\text{ K/km}$). As a result of this vertical structure, refractivity decreases steadily with height above ground level ($h$):

$$\frac{dN}{dh} \approx -39\text{ N-units/km} \quad \left(\text{or } \frac{dn}{dh} \approx -39 \times 10^{-9}\text{ m}^{-1}\right)$$

Because the refractive index decreases with altitude, the upper portion of an advancing electromagnetic wavefront travels slightly faster than the lower portion ($v_{\text{top}} > v_{\text{bottom}}$). This velocity differential causes the wave to continuously tilt downward toward the Earth's surface. The radius of curvature ($r$) of the refracted ray path is:

$$\frac{1}{r} = -\frac{dn}{dh} \approx 39 \times 10^{-9}\text{ m}^{-1} \implies r \approx 25,600\text{ km}$$

Because this downward ray curvature ($r \approx 25,600\text{ km}$) is less than the physical curvature of the Earth ($a \approx 6371\text{ km}$), the wave does not follow the planet indefinitely, but it bends around the curve far more than a straight optical ray.

To simplify RF link planning, engineers replace the curved ray path over a curved Earth with an equivalent straight-line ray propagating over a fictitious, enlarged Earth of effective radius ($a_e = k a$). The effective Earth radius factor ($k$) is given by:

$$k = \frac{1}{1 + a \frac{dn}{dh}} = \frac{1}{1 - 6371 \times 10^3 \times 39 \times 10^{-9}} \approx \frac{1}{1 - 0.2485} \approx \frac{4}{3} \approx 1.333$$

Derivation of the Radio Horizon Formula

From the geometry of a sphere of effective radius $a_e$, the distance ($d$) from an antenna at height $h$ above flat ground to the tangential horizon point is given by the Pythagorean theorem:

$$(a_e + h)^2 = a_e^2 + d^2 \implies a_e^2 + 2 a_e h + h^2 = a_e^2 + d^2$$

Since the antenna height $h$ is negligible compared to the planetary radius ($h \ll a_e$), the $h^2$ term can be omitted:

$$d \approx \sqrt{2 a_e h}$$

For the visual (optical) horizon over true Earth ($a = 6371\text{ km}$, where $h$ is in metres and $d$ is in kilometres):

$$d_{\text{optical}} \approx \sqrt{2 \times 6371 \times 10^{-3} \times h} \approx 3.57 \sqrt{h}\text{ (km)}$$

For the radio horizon under standard $k = 4/3$ atmospheric refraction ($a_e = 8495\text{ km}$):

$$d_{\text{radio}} \approx \sqrt{2 \times 8495 \times 10^{-3} \times h} = \sqrt{16.99 h} \approx 4.12 \sqrt{h}\text{ (km)}$$

When two stations at antenna heights $h_1$ and $h_2$ (metres above ground level) establish a line-of-sight path over smooth terrain, the maximum unobstructed radio path length ($d_{\text{LOS}}$) is the sum of their individual horizon distances:

$$d_{\text{LOS}} = d_1 + d_2 \approx 4.12 \times \left(\sqrt{h_1} + \sqrt{h_2}\right)\text{ km}$$

Comparing the two multipliers ($4.12 / 3.57 \approx 1.154$), standard atmospheric refraction extends the radio horizon approximately $15.4\%$ beyond the optical horizon.

Step-by-Step Worked Example: Repeater Line-of-Sight Distance

An amateur radio club installs a $147\text{ MHz}$ 2-meter VHF repeater on a hilltop mast with an antenna height of $h_1 = 64\text{ metres}$ above surrounding terrain. A mobile operator uses a quarter-wave whip antenna mounted on a vehicle roof at $h_2 = 2.25\text{ metres}$. Calculate:

1. The optical horizon distance between the two antennas.
2. The radio horizon distance under standard $k=4/3$ atmospheric refraction.
3. The additional communication distance gained purely from tropospheric refraction.

Step 1: Calculate the optical horizon distance:

$$d_{\text{optical}} = 3.57 \times (\sqrt{64} + \sqrt{2.25}) = 3.57 \times (8.0 + 1.5) = 3.57 \times 9.5 \approx 33.92\text{ km}$$

Step 2: Calculate the radio horizon distance:

$$d_{\text{radio}} = 4.12 \times (\sqrt{64} + \sqrt{2.25}) = 4.12 \times (8.0 + 1.5) = 4.12 \times 9.5 \approx 39.14\text{ km}$$

Step 3: Calculate the distance gained:

$$\Delta d = d_{\text{radio}} - d_{\text{optical}} = 39.14\text{ km} - 33.92\text{ km} = 5.22\text{ km}$$

Standard atmospheric refraction extends the reliable direct coverage footprint by more than $5.2\text{ km}$ in radius.

---

Tropospheric Ducting and Super-Refraction

Under non-standard weather conditions, the vertical profile of temperature, pressure, and humidity departs drastically from the standard lapse rate. These deviations alter the refractivity gradient ($dN/dh$), leading to four distinct refractive regimes:

1. Sub-refraction ($dN/dh > -39\text{ N-units/km}$, relative to this standard model): Downward bending is weaker than normal. At $dN/dh=0$ rays are straight; a positive gradient bends them upward. The radio horizon is shorter than in the standard atmosphere.
2. Standard Refraction ($dN/dh \approx -39\text{ N-units/km}$): Normal downward bending with $k = 4/3$.
3. Super-refraction ($-157 < dN/dh < -39\text{ N-units/km}$): Downward curvature is more pronounced than normal ($k > 4/3$), extending the radio horizon significantly.
4. Trapping or Ducting ($dN/dh < -157\text{ N-units/km}$): The downward radius of curvature of the wave becomes smaller than the radius of the Earth ($r < a$). The effective Earth radius factor $k$ becomes negative or infinite. The radio wave is trapped between atmospheric boundaries or between an atmospheric boundary and the ground.

                 Surface duct: schematic ray paths

       Strong fall of refractivity with height
       ----------------------------------------
          __        __        __
       __/  \______/  \______/  \___  guided rays
       ========================================
                    Sea / ground

       Rays curve within the refractive layer;
       confinement depends on frequency and launch angle.

Physical Mechanisms of Duct Formation

For trapping to occur ($dN/dh < -157\text{ N-units/km}$), the term $(P + 4810e/T)$ in the refractivity equation must decrease extremely rapidly with altitude. This is driven by two meteorological conditions acting individually or together:

• Temperature Inversion: Temperature increases with altitude ($dT/dh > 0$) rather than dropping, causing warm dry air to sit directly above cooler air.
• Steep Water Vapour Gradient: The humidity drops precipitously with altitude ($de/dh \ll 0$), such as directly above an ocean surface where a thin, saturated layer is capped by bone-dry continental air.

Three primary meteorological scenarios trigger ducting:

• Nocturnal Radiation Cooling: On clear, windless nights, the Earth's surface radiates heat rapidly into space, chilling the ground and the immediate surface air layer. Warmer air remains aloft, creating a shallow ground-based temperature inversion that forms after dusk and dissipates shortly after sunrise.
• High-Pressure Anticyclonic Subsidence: Large, slow-moving high-pressure systems feature descending air that warms adiabatically as it compresses. This creates an elevated subsidence inversion layer between $500\text{ and }2000\text{ metres}$ altitude, capable of forming vast elevated ducts spanning thousands of kilometres.
• Coastal Advection: Warm, dry air from hot desert or continental interiors blows out over cold ocean waters. The cold sea chills the lowest layer of air while evaporation charges it with moisture, establishing a permanent or semi-permanent maritime surface duct.

Spreading and Range Extension

In free space, power density from a given transmitter decreases as $1/d^2$, corresponding to a $20\text{ dB}$ increase in spreading loss per decade of distance.

An ideal duct that confines energy vertically can approach cylindrical spreading, with power density varying as $1/d$ and spreading loss increasing by $10\text{ dB}$ per decade. This illustrates why ducting can greatly enhance distant signals. A real atmospheric duct also leaks energy and has surface, absorption and coupling losses; its total path loss is not fixed by this ideal spreading law. Long maritime paths can open with modest stations, but transmitter power alone cannot predict received signal strength.

Frequency Limits and Duct Depth

An atmospheric duct resembles a waveguide, but its boundaries are refractive gradients rather than metal walls. Its ability to confine a wavelength depends on both its depth and the strength and shape of its refractivity profile. There is no universal depth threshold for each amateur band, and the transition to poorly confined propagation is gradual.

• A deeper or stronger duct can confine longer wavelengths than a shallow, weak duct.
• Shallow maritime ducts often favour UHF and microwave propagation.
• Antenna height and elevation angle determine how effectively a station couples energy into the duct.

Southern Australian Maritime Paths

The Great Australian Bight and Bass Strait provide useful examples of over-water paths on which inversions and humidity gradients can enhance VHF/UHF communication. Openings depend on the position and continuity of the refractive layer, antenna coupling and the station link budget. Listen for distant beacons and compare bands rather than assuming a particular range or signal strength from the weather chart alone.

---

Terrain Diffraction and Obstacle Clearance

When a radio wave encounters terrain, diffraction carries some energy into the geometric shadow. The Huygens-Fresnel principle describes this through contributions from the unobstructed wavefront. A sharp isolated ridge can be approximated by a knife edge. Rounded hills, multiple obstacles and the curved Earth require other diffraction models.

                 Single knife-edge geometry

                         summit
                           o
                          /|\
                         / |h\
       Tx o---------------+---------------o Rx
             <--- d1 --->  <--- d2 --->
             <--------- D=d1+d2 -------->

       h is positive above the direct Tx-Rx ray;
       it is negative when the summit is below it.

The Knife-Edge Diffraction Parameter

The attenuation caused by an obstacle depends on the dimensionless Fresnel-Kirchhoff diffraction parameter ($\nu$):

$$\nu = h \sqrt{\frac{2 (d_1 + d_2)}{\lambda d_1 d_2}} = h \sqrt{\frac{2 D}{\lambda d_1 d_2}}$$

where:

• $h$ is the clearance height of the obstacle apex above the direct line-of-sight ray path (defined as positive if the obstacle penetrates above the ray line, and negative if the apex is below the line),
• $d_1$ is the distance from the transmitter to the obstacle,
• $d_2$ is the distance from the obstacle to the receiver,
• $D = d_1 + d_2$ is the total path length,
• $\lambda$ is the radio wavelength.

When the obstacle apex lies precisely on the line-of-sight path between the two antennas ($h = 0$, grazing incidence):

$$\nu = 0 \implies \text{Diffraction loss is exactly } 6.0\text{ dB}$$

A grazing obstruction blocks half of the first Fresnel zone's electric field, halving the received voltage ($V / V_0 = 0.5$) and quartering the power, which corresponds to an unavoidable $6\text{ dB}$ obstacle loss.

Fresnel Zones and Clearance Criteria

To ensure that an RF path achieves free-space propagation without diffraction attenuation, the line of sight must not merely clear the geometric crest of intervening terrain—it must maintain adequate clearance across its Fresnel zones.

Fresnel zones are concentric, football-shaped (prolate spheroidal) boundaries surrounding the direct ray path. The $n$-th Fresnel zone boundary represents the locus of points where an indirect wave reflected from the boundary travels a distance that is $n$ half-wavelengths ($n \lambda / 2$) longer than the direct ray:

$$\Delta d = d_{\text{indirect}} - d_{\text{direct}} = n \frac{\lambda}{2}$$

For the crucial First Fresnel Zone ($n = 1$), the path length difference is $\lambda / 2$ ($180^\circ$ phase shift). The radius of the first Fresnel zone ($R_1$) at any point along the link is:

$$R_1 = \sqrt{\frac{\lambda d_1 d_2}{d_1 + d_2}} = 17.32 \sqrt{\frac{d_1 d_2}{f_{\text{GHz}} D_{\text{km}}}}\text{ (metres)}$$

where $d_1, d_2,$ and $D$ are in kilometres, and $f_{\text{GHz}}$ is the carrier frequency in GHz.

Step-by-Step Worked Example: First Fresnel Zone Radius Calculation

An amateur radio operator designs a $24\text{ km}$ point-to-point link on the $1.296\text{ GHz}$ ($23\text{ cm}$) band. A sharp mountain ridge crosses the path at a distance of $d_1 = 8\text{ km}$ from the transmitter and $d_2 = 16\text{ km}$ from the receiver ($D = 24\text{ km}$). Calculate:

1. The radius of the first Fresnel zone ($R_1$) at the mountain ridge.
2. The minimum vertical clearance required above the ridge crest to eliminate diffraction loss.

Step 1: Calculate $R_1$ using the formula:

$$R_1 = 17.32 \sqrt{\frac{d_1 d_2}{f_{\text{GHz}} D_{\text{km}}}} = 17.32 \sqrt{\frac{8 \times 16}{1.296 \times 24}} = 17.32 \sqrt{\frac{128}{31.104}}$$ $$R_1 = 17.32 \sqrt{4.1152} = 17.32 \times 2.0286 \approx 35.14\text{ metres}$$

Step 2: Calculate the required $60\%$ clearance:

$$\text{Clearance} = 0.60 \times R_1 = 0.60 \times 35.14\text{ m} \approx 21.08\text{ metres}$$

The direct visual line between the two antennas must pass at least $21.1\text{ metres}$ above the summit of the mountain ridge to prevent diffraction loss.

---

Rain Scatter and Tropospheric Scatter

Beyond direct line of sight and ducting, scattering can provide additional paths. Their usefulness depends on the scattering volume, station equipment and fading.

1. Rain Scatter (Precipitation Scatter)

Rain scatter is useful on microwave bands, especially $5.7\text{ GHz}$ and $10\text{ GHz}$, and is also possible at $24\text{ GHz}$. At $10\text{ GHz}$ the wavelength is about $30\text{ mm}$, larger than typical millimetre-sized raindrops. Scattering increases strongly with frequency in the Rayleigh region; at higher frequencies particle size and additional attenuation become increasingly important.

• Operating Technique: Stations do not point their dish or horn antennas toward each other. Instead, both operators aim their narrow antenna beams toward a common intense rain cell or thunderstorm located within their mutual line of sight, often $100\text{ to }400\text{ km}$ away.
• Doppler Broadening: Raindrops within a storm cell fall at differing terminal velocities and are buffeted by turbulent convective updrafts and downdrafts moving at $10\text{ to }30\text{ m/s}$. The reflections off these randomly moving droplets impose extreme Doppler broadening ($100\text{ to }500\text{ Hz}$) on the scattered signal.
• Audio Effect: CW and SSB can acquire a raspy sound from Doppler spreading. CW is often easier to copy, and strong signals can support narrowband FM. Digital decoding depends on the mode’s tolerance of the spread and fading; readability is not guaranteed.

2. Tropospheric Scatter (Tropo-scatter)

Small irregularities in tropospheric refractive index scatter a small part of the transmitted energy forward. Both stations aim directional antennas toward a common volume beyond their horizons. The additional loss depends on frequency, scattering angle, refractivity and antenna coupling.

• Availability: The mechanism is continuously present, unlike an occasional duct opening. Received signals still undergo slow atmospheric variations and rapid fading, so continuous readability requires adequate link margin.
• Range: Paths of a few hundred kilometres, for example $200\text{ to }600\text{ km}$, are useful illustrations. Terrain shielding, antenna horizon angles and the link budget determine whether a particular path works.
• Equipment: Directional antennas, low feeder loss, low receiver noise and narrowband modes such as CW or Q65 improve the link budget. Use sufficient lawful power: in the relevant Advanced bands the class licence permits $400\text{ W}$ pX for C3F/J3E/R3E and $120\text{ W}$ mean power for other emissions, including CW and Q65.

---

Propagation Mode Comparison