So if you know any two of those, you solve for the third — Page 379, Lesson 382

Let’s start with the very first one, Q1, because it sets the tone for how we’ll think about these. The question is about the relationship between frequency and wavelength, and the hint is simply: use c = f λ. That’s the universal wave equation. c is the speed of light, roughly 300,000,000 metres per second. f is the frequency in hertz, and λ, the Greek letter lambda, is the wavelength in metres. So if you know any two of those, you solve for the third. In radio navigation, this is your bread and butter — every time you’re told a frequency, you can immediately work out the wavelength, and vice versa. Keep that equation in your pocket. Now Q5. This one is about line of sight, and the formula is: Range in nautical miles equals 1.23 times the square root of the transmitter height plus the square root of the receiver height. Written out: Range (NM) = 1.23 (√hTX + √hRX), with heights in feet. The key idea here is that VHF and UHF signals — the ones used for things like VOR and DME — travel in straight lines. They don’t bend over the horizon. So the maximum range you can get is limited by the curvature of the Earth. The higher your antenna, the further you can see, and the formula captures that. The 1.23 is a constant that converts feet and the Earth’s radius into nautical miles. And note it’s the sum of the square roots of both heights — transmitter and receiver — because both ends contribute to the line of sight. Q13 is the same formula again, just applied. The question tells you the maximum range at which reception can be achieved is 195 nautical miles. So you’re using the line of sight formula in reverse, or plugging in numbers to verify. The point is: when you see a range question, your first instinct should be line of sight. Q14 is a conceptual one. It says: the phase difference is the bearing of the aircraft from the beacon — that’s the radial. Let me unpack that. In a VOR, the ground station transmits a signal whose phase varies with direction. The aircraft receiver measures the phase difference between a reference signal and a variable signal. That phase difference, in degrees, directly corresponds to the radial — the bearing of the aircraft from the beacon, measured clockwise from magnetic north. So if the phase difference is 90 degrees, you’re on the 090 radial. That’s the whole principle of VOR. Q15 and Q16 just say: draw a diagram. These are geometry questions, likely about bearings or tracks, and the advice is to sketch it out. When you draw the triangle, the angles and distances become obvious. Don’t try to do it in your head — put pencil to paper. Q18 is a height calculation. The formula is: Height = Glide path angle × range × 100 feet. So if you’re on a 3-degree glide path and you’re 5 nautical miles from the threshold, your height is 3 × 5 × 100 = 1500 feet. The 100 is the conversion factor that turns the product of degrees and nautical miles into feet. This is the classic ILS descent calculation — it tells you what altitude you should be at for a given distance. Q27 — there’s no hint text, just the question number. That suggests it’s a diagram-based or formula-based question where the answer is self-evident once you see the figure. I’d treat it as a reminder to apply whatever principle was just covered. Q36 is about Mode C, the altitude-reporting transponder mode. It says: Mode C increments in 100-foot steps. That means the altitude information is quantised — it reports in discrete 100-foot intervals, not continuously. So if you’re at 12,350 feet, Mode C will report 12,300 or 12,400, depending on rounding. This matters for separation — you can’t rely on Mode C for precision below 100 feet. Q37 is a frequency allocation question. It says: 1262 MHz is outside the allocated band for DME. DME operates in the 960 to 1215 MHz band, so 1262 is too high. This is a classic trap — they give you a plausible-looking frequency, but it’s out of range. Know your bands cold. Q39 just says: Pythagoras. So this is a right-angled triangle problem — likely a ground speed or distance calculation where you have two sides and need the third. Remember: a² + b² = c². In navigation, this often comes up when you have a crosswind component or a slant range. Q54 is about GPS. It says: the 50 Hz modulation passes the Nav and System Data message. The PRN codes provide a timing function and SV identification. Let me break that down. Each GPS satellite transmits on the same frequency, but with a unique Pseudo-Random Noise code — that’s the PRN. That code does two things: it lets the receiver identify which satellite is which, and it provides the timing reference. The 50 Hz data stream, modulated on top, carries the navigation message — the satellite’s position, clock corrections, and system health. So you have three layers: the carrier, the PRN code for timing and ID, and the 50 Hz data for the actual navigation information. Q57 says: the range displayed is to the waypoint. This is about an FMS or GPS display — when you’re navigating to a waypoint, the distance shown is the distance to that waypoint, not to the next one or to the destination. Simple but easy to confuse. Q62 says: remember the PLAN display is orientated to TRUE north. This is a classic trap. On a moving map or plan view, the map is drawn with true north at the top, not magnetic north. So when you read a bearing off the plan display, it’s a true bearing, and you’ll need to apply variation to convert to magnetic for your compass or heading. Forget that, and you’ll be off by the local variation. Finally, at the bottom there’s a fragment: Time interval / 2 × 6.17 = Range in NM. That looks like a DME slant range or a radar range calculation. The 6.17 is the conversion factor — it’s the time, in microseconds, for a pulse to travel one nautical mile and come back, divided by 2 because it’s a round trip. So if you measure a time interval, you divide by 2 to get the one-way time, multiply by 6.17, and you get the range in nautical miles. That’s the principle behind DME ranging — it measures the time for a pulse to go out and come back. So, the big themes across these answers: the wave equation, line of sight, phase difference as bearing, glide path height, Mode C increments, frequency bands, Pythagoras, GPS signal structure, and the true north orientation on displays. Each one is a tool you’ll use repeatedly.