Question E3B07:
What effect does lowering a signal’s transmitted elevation angle have on ionospheric HF skip propagation?  

Choices:  
A: Faraday rotation becomes stronger  
B: The MUF decreases  
C: The distance covered by each hop increases  
D: The critical frequency increases    

Correct Answer is  C: The distance covered by each hop increases  

Correct answer: C — The distance covered by each hop increases

Why C is correct
- Lowering the transmitted elevation angle (smaller angle above the horizon) makes the skywave path shallower. For a given ionospheric layer height, a shallower incidence angle means each “bounce” lands farther away.
- A useful geometry approximation for a single-hop path with virtual reflection height h and elevation angle ε is:
  $$D \approx 2h \cot(\varepsilon)$$
  Since $\cot(\varepsilon)$ increases as $\varepsilon$ decreases, $D$ increases as you lower the elevation angle.

Quick numeric example (h = 300 km):
- At $\varepsilon = 30^\circ$: $D \approx 2 \cdot 300 \cdot \cot 30^\circ \approx 600 \cdot 1.732 \approx 1040$ km
- At $\varepsilon = 10^\circ$: $D \approx 2 \cdot 300 \cdot \cot 10^\circ \approx 600 \cdot 5.671 \approx 3400$ km

Why the other choices are incorrect
- A: Faraday rotation becomes stronger  
  Faraday rotation depends on electron density, geomagnetic field, path length, and frequency. Changing takeoff angle isn’t the primary control; the notable, predictable effect of lowering elevation angle for HF skip is increased hop distance, not a systematic increase in Faraday rotation.

- B: The MUF decreases  
  Opposite is true. By the secant law,
  $$\text{MUF} \approx f_\text{critical} \sec(\chi)$$
  where $\chi$ is the incidence angle relative to vertical. Lower elevation angles correspond to larger $\chi$ and hence larger $\sec(\chi)$, so MUF increases. A common flat-path rule of thumb is:
  $$\text{MUF} \approx \frac{f_\text{critical}}{\sin(\varepsilon)}$$
  so as $\varepsilon$ decreases, $\sin(\varepsilon)$ decreases and MUF increases.

- D: The critical frequency increases  
  The critical frequency $f_\text{critical}$ is a property of the ionospheric layer (electron density), not your antenna’s elevation angle. It does not change with takeoff angle. A common relation is:
  $$f_\text{critical} \approx 9 \sqrt{N_\text{max}} \text{ MHz}$$
  where $N_\text{max}$ is the peak electron density.

Rules of thumb
- Lower takeoff angles (about 5–15°) favor long-haul DX due to longer hop distances and higher usable frequencies.
- Higher takeoff angles (e.g., 60–90°) favor NVIS/regional coverage with short skip distances.
- Skip distance grows as $\varepsilon$ decreases: $D \propto \cot(\varepsilon)$.