Let’s pick this up right where the worked answers left off — Page 565, Lesson 562

Let’s pick this up right where the worked answers left off. I’m going to walk you through the reasoning behind each of these answers, because they’re not just about getting the right letter — they’re about the method you’ll use in the exam and in the aircraft. First, the change-of-longitude question. The key line is: change of longitude × 0.80. That’s a conversion factor. At the latitude in question, 1 minute of longitude is not 1 nautical mile — it’s only 0.80 of a nautical mile. So if you have a change of longitude of 15°, you multiply by 0.80 to get the equivalent distance in nautical miles. And the second part: 15° eastwards from 004°W is 011°E. Let’s check that. You’re at 4° west. Going east 15° means you subtract 15 from 4 — but you cross the Greenwich meridian. From 4°W to 0° is 4°, then you have 11° more to go, so you end up at 11°E. That’s the arithmetic of crossing the prime meridian. Next, the resolution of a track into latitude and longitude change. You’re flying 70 NM along a track of 225°. Track 225 is southwest — exactly halfway between south (180) and west (270), so it’s 45° from each. You resolve it: 70 sin 45 gives the south component, 70 cos 45 gives the west component. Both come out to 49 NM. So you’ve moved 49 NM south and 49 NM west. Now here’s the clever bit: because you’re at the Equator, 1 minute of longitude equals 1 nautical mile — so you can take 49 NM west directly as 49 minutes of longitude change, no departure formula needed. The final position is therefore 49 minutes south and 48 minutes west of the original. Note the slight difference — 49 west but 48 in the answer — that’s just rounding in the original working. Now the Earth-shape question. The language is the only trap. The polar and equatorial diameters are in the ratio 296:297. The semi-major axis is half the equatorial diameter — that’s the Earth’s radius at the equator, given as 6378.4 km. The semi-minor axis is half the polar diameter — that’s the polar radius, which is what we want. So you set up the proportion: Polar radius / 6378.4 = 296 / 297. Solving gives 6356.9 km. The alternative method: remember the polar diameter is 43 km shorter than the equatorial diameter, so the radius is 21.5 km shorter. Subtract: 6378.4 – 21.5 = 6356.9 km. Either way, same answer. Then we have a combined departure and scale problem. Departure = change of longitude × cos latitude. Here, change of longitude is 10°, converted to minutes by multiplying by 60, and cos latitude is 0.7071 — that’s cos 45°. So departure = 10 × 60 × 0.7071 = 424.3 NM. That’s the actual distance on the Earth’s surface. Now for scale: Scale = Chart Length / Earth Distance. The chart length is 14 cm. The Earth distance is 424.3 NM, converted to centimetres: multiply by 1852 (metres per nautical mile) and then by 100 (centimetres per metre). That gives a scale of approximately 1 in 5.6 million, which matches answer (d). Finally, the descent profile question. You’re at FL370 and need to descend to FL80. The difference is 29,000 feet. At 1800 feet per minute, that’s 16.1 minutes of descent. Your mean ground speed in the descent is 232 knots, so in 16.1 minutes you cover 62 NM during the descent. Your total distance to run is 185 NM. If 62 of those are the descent, then the distance at high level to the top of descent is 185 – 62 = 123 NM. At your level ground speed of 320 knots, 123 NM takes 23 minutes. Add that to your current time of 0422, and you get 0445 as the ETA for top of descent. And the last one — the NDB relative bearing question. True heading is 140, true track is 150. At point A, the NDB is 35° left of the nose, so the true bearing to the NDB is 140 – 35 = 105°. At point B, the NDB is 80° left of the nose, so the true bearing is 140 – 80 = 060°. That’s the classic relative-bearing-to-true-bearing conversion: true bearing = true heading + relative bearing, where left is negative. That’s the full set of methods behind those answers. Each one is a building block — the departure formula, the scale calculation, the descent time/distance, and the bearing conversion — and you’ll use all of them again in the exam.