
Let’s pick this up right where the method leaves off. We’ve already got the LMT of sunrise or sunset from the Air Almanac, and we’ve converted it to UTC or Standard Time using the arc-to-time and zone-time methods from the previous chapter. Now I want to show you how that plays out in real problems, and there’s one crucial rule that governs all of them: interpolation is only required to the nearest minute. If the date you need isn’t one of the dates listed in the Almanac, you interpolate, but you never go beyond the nearest minute. That’s a hard limit.
So the whole idea is this: sunrise and sunset problems are really ordinary time problems. The starting point is always the LMT — Local Mean Time — of sunrise or sunset at a given position. You extract that LMT from the Air Almanac, then convert it to UTC or Standard Time as required. Let me walk you through three worked examples, because they show you every step and every trap.
Example 1: Give the Standard Time of sunset at Innsbruck, Austria, at 47°15’N, 011°20’E, on 5th September.
First, from the Air Almanac, sunset at 47°15’N on the 5th is 18 hours 33 minutes LMT. Now I convert that LMT to UTC. The longitude is 11°20’E. Arc to time: 11 degrees is 44 minutes, and 20 minutes of arc is 1 minute 20 seconds of time — but we round to the nearest minute, so that’s 45 minutes total. Since Innsbruck is east of Greenwich, local time is ahead of UTC, so I subtract the arc-to-time: 18:33 LMT minus 45 minutes gives 17:48 UTC on the 5th.
Now I convert UTC to Standard Time. Austria is in List 1, which is UTC plus 1 hour. So 17:48 UTC plus 1 hour gives 18:48 Standard Time on the 5th. That’s the answer: sunset at Innsbruck is at 18:48 Standard Time.
Example 2: What is the Standard Time of sunrise at Keflavik, Iceland, at 64°00’N, 22°30’W, on 14th October? And note — Summer Time is not being kept.
From the Air Almanac, sunrise at 64°N on the 14th is 06:45 LMT. Longitude is 22°30’W. Arc to time: 22 degrees is 1 hour 28 minutes, and 30 minutes of arc is 2 minutes of time, so that’s 1 hour 30 minutes total. Keflavik is west of Greenwich, so local time is behind UTC, meaning I add the arc-to-time to get from LMT to UTC: 06:45 plus 1 hour 30 minutes gives 08:15 UTC on the 14th.
Now Iceland is in List 2, which is UTC plus 0 hours — that’s UTC itself. So Standard Time is 08:15 on the 14th. That’s the answer.
Example 3 is the one that matters operationally. An aircraft lands at Goose, Labrador, Canada, at 53°20’N, 060°20’W, at 18:41 Standard Time on 20th September. Is it a day or night landing? Ignore Summer Time.
Here’s the key definition you must know cold: the Air Navigation Order defines the period of night flying as from 30 minutes after sunset to 30 minutes before sunrise at the surface. Note the qualifier — this is not necessarily the same as when it gets dark. It’s a legal definition, not a physical one.
So first, from the Air Almanac, sunset at Goose on the 20th is 18:05 LMT. Longitude is 060°20’W. Arc to time: 60 degrees is 4 hours, and 20 minutes of arc is 1 minute 20 seconds, rounded to 1 minute. That gives 4 hours 1 minute. Goose is west, so I add: 18:05 LMT plus 4 hours 1 minute gives 22:06 UTC of sunset at Goose on the 20th.
Now convert to Labrador Standard Time. Labrador is in List 3, which is UTC minus 4 hours. So 22:06 UTC minus 4 hours gives 18:06 Standard Time on the 20th. That’s sunset at Goose in Labrador Standard Time.
The aircraft lands at 18:41 Standard Time. Sunset is at 18:06. So the landing is 35 minutes after sunset. Since night flying starts 30 minutes after sunset, and 35 minutes is past that, this is a night landing according to the ANO definition.
Notice the structure of every problem: LMT from the Almanac, arc-to-time conversion to UTC, then zone-time conversion to Standard Time. And always keep the sign straight — east means local time is ahead, so you subtract to get UTC; west means local time is behind, so you add. That’s the whole method, and these three examples cover every variation you’ll see.
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