Let’s pick this up right where the calculation left off — Page 560, Lesson 556

Let’s pick this up right where the calculation left off. You’ve got a departure of 3000 km, and you’re comparing what happens at 27°N versus at the Equator. The key insight is that a given east–west distance on the Earth’s surface corresponds to a larger change of longitude at higher latitudes. At the Equator, 3000 km covers fewer degrees of longitude; at 27°N, the same 3000 km covers more degrees of longitude. So the answer must be at 27°N, but west of the 170°W meridian — because you’re moving west from that meridian, and the longitude change is bigger up at 27°N. Only one option fits that. Now, question 59 is a plotting question — you solve it by measuring directly on the chart. No calculation shortcut; just measure. Question 60: you have 120 NM of horizontal distance to travel at 288 knots. That gives you 25 minutes of flying time. You need to lose 24,000 feet in those 25 minutes, which works out to 960 feet per minute rate of descent. Question 61: you need to lose 6,500 feet at 1,000 feet per minute. That takes 6.5 minutes. At a ground speed of 156 knots, flying for 6.5 minutes covers 16.9 NM. But that’s 6 NM short of the DME, so you add them: total distance to start the descent is 22.9 NM. Question 63: a 9-degree change of latitude. You could compute 540 NM and convert to km, but there’s a faster way. 90 degrees of latitude change equals 10,000 km, so 9 degrees is 1,000 km. Question 64 — this is the Mercator chart one. On a Mercator chart, meridians are drawn as parallel lines. So the total length of the 53°N parallel is 133 cm, and the 30°S parallel is also 133 cm. Now find the departure from 180°E to 180°W at 30°S. Departure = change of longitude × cos latitude. That’s 360 × 60 × cos 30 = 18,706 NM. Then it’s a simple scale problem: chart length 133 cm, Earth distance 18,706 NM, giving a scale of approximately 1:26 million. Question 65: the Sun would be at its zenith. Question 66 — compass deviation. Find true heading normally, then apply variation and deviation. First check you’ve balanced the drift. When deviation is given as East or West, use the rule: deviation East, compass least. But deviation — not variation, only deviation — is sometimes quoted as plus or minus. In that case, deviation is what you apply to the compass to get magnetic, not the other way round. So East deviation is plus, West deviation is minus. Question 69: 3 × 444 is 1,332 km. Divide by 1.852 to get 719.2 NM, approximated to 720, giving a 12° change of latitude, which takes you to 02°S. Question 70: the average great circle track is the rhumb line track. Question 71: distance from Equator to a Pole is 5,400 NM — that’s 90° × 60 NM per degree. So Earth’s circumference is 4 × 5,400 = 21,600 NM. Alternatively, 360° × 60. Question 73: the rhumb line track from 70°S to 70°S is along a parallel of latitude, so it’s 090°(T). But the great circle track cuts the corner and takes the shortest route, which is south of 090°(T). Draw it on a Lambert’s projection or try it on a globe. That’s the whole set. You’ve got the reasoning for each one now.