First, Q1 — Page 379, Lesson 382

We're in the answers section now, and I want to walk you through the explanations for the selected questions. These are the key worked examples that tie together the whole chapter, so let's go through them one by one. First, Q1. The hint is simply: use c = f λ. That's the fundamental wave equation. c is the speed of light, f is frequency, and λ is wavelength. If you're given any two of these, you can find the third. This is the backbone of all radio navigation — every frequency we use has a corresponding wavelength, and that relationship governs how the signal behaves. Now Q5 and Q13 both use the line of sight formula. This is critical for VHF and UHF systems. The formula is: Range in nautical miles equals 1.23 times the sum of the square roots of the transmitter height and receiver height, where both heights are in feet. So it's Range (NM) = 1.23 (√hTX + √hRX). The transmitter height is hTX, the receiver height is hRX. In Q13, applying that formula gives a maximum reception range of 195 NM. That's the absolute limit — beyond that, the Earth's curvature blocks the signal, no matter how powerful the transmitter is. Q14 is about phase difference. The key point here is that the phase difference is the bearing of the aircraft from the beacon — that's the radial. So when you measure phase, you're directly measuring your direction relative to the station. Q15 and Q16 just say: draw a diagram! This is a reminder that for geometry-based questions, sketching the situation makes the relationship obvious. Q18 gives you a height formula: Height = Glide path angle × range × 100 ft. So if you're on a 3-degree glide path at a range of 5 NM, your height is 3 × 5 × 100 = 1500 feet. This is the standard way to compute your height on approach. Q27 has no explanation listed — it's just there as a question reference. Q36 is about Mode C. The key fact: Mode C increments in 100 ft steps. So the altitude encoder reports altitude in discrete 100-foot increments, not continuously. Q37: 1262 MHz is outside the allocated band for DME. That's a frequency check — DME operates in a specific band, and 1262 MHz doesn't fall within it. Q39 just says: Pythagoras! That's the hint for a right-angled triangle problem — likely resolving a vector into components or finding a resultant. Q54 is about GPS signal structure. The 50 Hz modulation passes the Nav and System Data message. That's the slow data stream. The PRN codes — Pseudo-Random Noise codes — provide two functions: a timing function and SV identification. So each satellite has a unique PRN code that lets the receiver identify which satellite it's hearing and use the signal for precise timing. Q57: the range displayed is to the waypoint. That's a reminder about what distance readout you're actually looking at. Q62: remember the PLAN display is orientated to TRUE north. So when you're looking at a plan view on the navigation display, the top of the screen is true north, not magnetic. And finally, at the bottom, there's a time interval formula: Time interval = 2 × 6.17 / Range in NM. This is the classic DME round-trip time calculation — the signal goes out and comes back, hence the factor of 2. The 6.17 is the time in microseconds per nautical mile for the round trip. So if you know the range, you can compute the total time delay. That's the full set of explanations. These are the worked examples that show you how to apply each formula and concept from the chapter.