The conversion angle will be 42 degrees, so this makes the initial great… — Page 560, Lesson 558

Let’s pick this up right where the numbers start making sense. I’m going to walk you through the tail end of this answer key, because there are a few real navigation gems buried in here — the conversion angle, the departure formula, and a time-zone calculation. First, the opening line: “The conversion angle will be 42 degrees, so this makes the initial great circle track 132(T).” This is the classic great-circle-versus-rhumb-line relationship. The conversion angle is half the change of longitude times the sine of the mean latitude — that’s the formula you’ll see again and again: conversion angle = ½ × change of longitude × sin(mean latitude). Here, the conversion angle comes out to 42 degrees, and you apply it to the rhumb line track to get the initial great circle track. So if the rhumb line track is 90(T), adding the 42-degree conversion angle gives you 132(T) as the initial great circle track. The key idea: a great circle starts off at a different angle than the rhumb line, and that difference is the conversion angle. Now, question 77 — this is a latitude and longitude reasoning problem. The statement is: “6°S is a greater value of latitude than 4°N.” Think about it — latitude is measured from the Equator, so 6°S is further from the Equator than 4°N. For a given departure — that is, a given east-west distance flown — a greater latitude means a greater change of longitude eastwards, because the meridians converge as you move away from the Equator. So when you fly east at 6°S, you cover more degrees of longitude than you would at 4°N. Then, when you come back to 4°N, the same 600 NM does not take you so far westwards, because at the lower latitude the meridians are further apart. So you finish up east of where you started. That’s the whole logic — greater latitude, greater longitude change for the same distance. Then we hit the fuel conversion problem, question 85. This is a classic unit conversion. The rule: there are 5 imperial gallons to 6 US gallons, and the imperial gallon to litres conversion is 4.55. Then you multiply by 0.78 to get the weight in kilograms. So the calculation is: 380 × 5/6 × 4.55 × 0.78 = 1123.85. Let me unpack that. 380 is the quantity in US gallons. Multiply by 5/6 to convert US gallons to imperial gallons. Then multiply by 4.55 to convert imperial gallons to litres. Then multiply by 0.78 — that’s the fuel density in kilograms per litre. The result, 1123.85, is the weight in kilograms. The note says it’s probably easier with the CRP-5 — that’s the flight computer — but the arithmetic is there if you want to do it longhand. Question 86 is a simple time-and-distance problem. Distance still remaining = 475 – 190 = 285 NM. Time to go = 1130 – 1040 = 50 minutes. So you’ve got 285 NM to cover in 50 minutes. That’s the setup — you’d then compute groundspeed required, but the answer just gives you the two components. Question 87 is a plotting question — solve by measurement on the chart. There’s a note about “both DME distances decreasing,” and the explanation is given in the Plotting chapter. I’ll flag that for you — DME, Distance Measuring Equipment, gives you slant range from the station. If both DME distances are decreasing, you’re flying toward both stations, which is a specific geometry you’ll solve on the chart. Question 88 — this is about Jeppesen conventions differing slightly from the ICAO ones. The key is given in the introduction to the Jeppesen Student Pilots’ Manual. So if you’re using Jeppesen charts, be aware their conventions aren’t identical to ICAO’s — always check the manual’s introduction. Questions 89, 90, 91 are all plotting questions — solve by measurement on the chart. No calculation to walk through; it’s pure chart work. Question 92 — this is a great-circle versus rhumb line comparison. The mean great circle is the same as the rhumb line track. The question asks which pairs of latitudes will give the greatest difference between great circle and rhumb line track — that is, which gives the greatest conversion angle. And the formula is: conversion angle = ½ × change of longitude × sin(mean latitude). So the greatest conversion angle comes from the greatest change of longitude and the greatest mean latitude — because sine increases with latitude up to 90 degrees. So you’re looking for the pair with the largest ch.long and the largest mean lat. Then there’s a note about temperature: “temperature normally decreases with increasing altitude. This means that the speed of sound will decrease so, for a given TAS, the Mach No. will increase, giving an additional effect.” Let me unpack that. TAS is True Airspeed — the actual speed of the aircraft through the air. Mach Number is the ratio of TAS to the speed of sound. As you climb, temperature drops, so the speed of sound drops. If your TAS stays the same, then dividing by a smaller speed of sound gives a larger Mach Number. So at altitude, for the same TAS, you’re flying at a higher Mach number. That’s the additional effect. Question 110 — this is a variation and bearing problem. Apply 17°W variation to 120°(M) to get 103°(T) heading. Remember the rule: variation west, magnetic best — so you subtract west variation from magnetic to get true. 120°M minus 17°W = 103°T. Then the island is 15°(T) to the left, which makes the true bearing TO the island 088°(T). So you take your true heading of 103° and subtract 15° because it’s to the left — 103 – 15 = 88°T. Question 111 — this is a departure problem. In one hour, the aircraft covers 360 NM. The departure formula is: Departure = change of longitude (minutes) × cosine latitude. So 360 NM = change of longitude (minutes) × cosine 60, and cosine 60 is 0.5. So change of longitude = 720 minutes. Then at the Equator, 720 minutes = 720 NM, which also has to be covered in one hour. So the groundspeed at the Equator would be 720 knots. The key relationship: departure is the east-west distance, and it equals the change of longitude in minutes times the cosine of the latitude. Finally, question 112 — this is a time-zone problem, and it uses a table. Standard Time at Kuwait: today at 0700 ST. The rule is “STD (long. east, UTC least)” — that means for longitudes east, UTC is less than standard time. So you subtract 3 hours to get UTC: 0700 ST minus 3 hours = 0400 UTC today. Then Standard Time at Algeria: today at 0500 ST. Same rule — Algeria is also east, so UTC is less. You add 1 hour to get UTC: 0500 ST plus 1 hour = 0600 UTC. Wait, let me re-read that. The table shows Algeria at 0500 ST, and the adjustment is +1 hour to get UTC. So 0500 ST plus 1 hour = 0600 UTC. The key principle: standard time zones east of Greenwich have UTC less than local standard time — you subtract the zone offset to get UTC. So there you have it — the conversion angle, the departure formula, the fuel conversion, and the time-zone arithmetic. These are the core navigation relationships you’ll use over and over.