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► 5 HR 20 MIN 20 SEC corresponds to a longitude difference of:
Answer: 80° 05 minutes
15° per hour
15 x (5° 20' 20") = 80° 5'
► How much time does it take for the Mean sun to move from meridian 14515E to meridian 02345W?
Answer: 11 hours 16 minutes
Difference in longitude = 169°
169° : 15°/H = 11H 16MIN
► What is the local mean time, position 65° 25 minutes N 123° 45 minutes W at 2200 UTC?
Answer: 1345
2200 UTC = at the prime meridian (0°E/W)
The position is 123°45' to the WEST of the prime meridian (West = Lags behind)
(123°45') : 15 = 08:15
The position is 08:15 behind 22:00 UTC
22:00 - 08:15 = 13:45
► Refer to the table: When it is 1000 Standard Time in Kuwait, the standard time in Algeria is:
Answer: 0800
1: Convert Kuwait ST into UTC using the table (it states that we should subtract 3H from standard time so 0700 UTC)
2: Convert UTC to ST Algeria using the table (it states that we should add 1H to UTC for standard time so 0800ST Algeria)
► Refer to the table: An aircraft takes off from Guam at 2300 Standard Time on 30 April local date. After a flight of 11 HR 15 MIN it lands at Los Angeles (California). What is the standard time and local date of arrival (assume summer time rules apply)?
Answer 1715 on 30 April
1: Guam 23:00 ST 30/04 = 13:00 UTC 30/04 (using the table)
2: Add the flight time to this to get landing time in LA at 24:15 UTC 30/04
3: In the table it states that LA is UTC-8 = 16:15 ST 30/04
4: Don't forget that summer time rules apply!! (+1 H) = 17:15 ST 30/04
► On the 27th of February, at 52°S and 040°E, the sunrise is at 0243 UTC. On the same day, at 52°S and 035°W, the sunrise is at:
Answer: 0743 UTC
Difference in longitude = 75°
75° : 15°/H = 5H
Position 2 is to the West so lags behind = 0243 + 0500 = 0743 UTC
(Position 2 will get the sunrise on a later time than position 1)
► On 4 February the Air Almanac lists 1941 as the time of sunset at 5000S. An observer registers sunset at 2113 UTC this day. What is the observers position?
Answer: 5000S 02300W
Difference in time = 01:32
01:32 x 15°/H = 23°
The observer will have a later sunset so position is to the West (23°W)
► For 1 February the Air almanac lists the following data: Latitude: 6600N Morning civil twilight: 0756 Sunrise: 0900 Sunset: 1528 Evening civil twilight: 1632. The duration of morning twilight at 6600N is:
Answer: 1 hour 4 minutes and starts at 0756 LMT
Morning civil twilight (0756) to Sunrise (0900) = 01:04
► When the time is 20:00 UTC, it is:
Answer: 14:00 LMT at 90°West
90° : 15°/H = 6H
Position is West so lags behind UTC = 20:00 - 06:00 = 14:00
(This is the only option out of the 4 where it gives the correct result)
► What is the standard time in Hawaii when it is 06:00 ST on the 16th July in Queensland, Australia?
Answer: 10:00 ST 15 July
Queensland = ST - 10 to get UTC
Because we can not do 06:00 - 10:00 we add 24:00 and subtract 1 day:
30:00 ST 15/07 - 10:00 = 20:00 15/07 UTC
Hawaii ST = UTC-10
Hawaii ST = 10:00 15/07
► The local mean time at longitude 095°20minW, at 0000 UTC, is:
Answer: 17:38:40 Previous day
Position is 95°20' to the WEST of the prime meridian
(95°20') : 15°/H = 06:21:20 BEHIND UTC
24:00 (so that we can use it in calculations we added 24H and thus it will be -1 day)
24:00 - 06:21:20 = 17:38:40
► The SR/SS table for the 23rd of February at latitude 40°N gives:
SR = 06:44
SS = 17:44
At 12:00 Central European Time (UTC+1) at 40°N:
Answer: The sun rises at 64°W
12:00 CET = 11:00 UTC
Difference between 11:00UTC and 06:44LMT = 04:16
04:16 x 15°/H = 64°
► Mua, Tonga Islands, is situated at (21° 11minS, 175° 07minW). In the Air Almanac the standard time of Tonga Islands is listed as UTC + 13. For August 21st the sunrise table in the air almanac shows:
20°S: 06:18
30°S: 06:28
What is the standard time of sunrise at Mua
Answer: 06:59 on August 22nd
Inter polar between 20°S and 30°S to find the time of sunrise for 21°21'S = 06:19 LMT
The longitude is 175°07'W so (: 15°/h) = 11:40
Position is to the West so sunrise will occur LATER = 06:19 + 11:40 = 17:59UTC
Mua ST is listed as UTC+13
Mua ST = 17:59 + 13:00 = 30:59 ST 21/08
This is of course not possible so remove 24:00 and add 1 day = 06:59 ST 22/08
► An aircraft follows a radial to a VOR/DME station.
At 10:00 the DME reads 120 NM.
At 10:03 the DME reads 105 NM.
The estimated time overhead the VOR/DME station is:
Answer: 10:24:00
First find the groundspeed that the aircraft has (we covered 15NM in 3 minutes)
15 NM : 3 min = 300kt
Now use your groundspeed and the total distance to find the total time it takes
120 NM : 300kt = 24 min
Now add the flight time to your start time
10:00 + 00:24 = 10:24
► An aircraft departs from Schiphol airport and flies to Santa Cruz in Bolivia (South America) via Miami in Florida. The departure time (off blocks) is 07:45 ST at the 10th of november. Taxi time before take off at Schiphol is 25 minutes. The flight time to Miami Over the Atlantic Ocean is 09:20. The total taxi time in Miami to and from the gate is 25 minutes. The time spend at the gate is 02:40. From Miami to Santa Cruz the airborne time is 06:30. Calculate the time and date of touch down in Santa Cruz in ST Bolivia if the difference between ST and UTC is 5 hours.
Answer: 21:05 10th Nov.
Schiphol off blocks time = 07:45 ST , convert this to UTC (Schiphol is UTC+1)
Schiphol off blocks time = 06:45 UTC
(the question does not state the difference between UTC and ST as being 1h so remember it for the exams should it come up!
)
Now you simply add up all the times in the order you read them in;
Off block = 06:45 UTC
+ 00:25 Taxi
= 07:10 UTC (take off time Schiphol)
+ 09:20 Flight time to Miami
= 16:30 UTC (landing time Miami)
+ 00:25 Taxi (both to and from Gate)
= 16:55 UTC
+ 02:40 Spend at Gate
= 19:35 UTC (take off time Miami)
+ 06:30 Flight time to Santa Cruz
= 26:05 UTC (landing in Santa Cruz)
Now convert to ST (given as "UTC-5") = 21:05 ST 10 Nov
Answer: 80° 05 minutes
15° per hour
15 x (5° 20' 20") = 80° 5'
► How much time does it take for the Mean sun to move from meridian 14515E to meridian 02345W?
Answer: 11 hours 16 minutes
Difference in longitude = 169°
169° : 15°/H = 11H 16MIN
► What is the local mean time, position 65° 25 minutes N 123° 45 minutes W at 2200 UTC?
Answer: 1345
2200 UTC = at the prime meridian (0°E/W)
The position is 123°45' to the WEST of the prime meridian (West = Lags behind)
(123°45') : 15 = 08:15
The position is 08:15 behind 22:00 UTC
22:00 - 08:15 = 13:45
► Refer to the table: When it is 1000 Standard Time in Kuwait, the standard time in Algeria is:
Answer: 0800
1: Convert Kuwait ST into UTC using the table (it states that we should subtract 3H from standard time so 0700 UTC)
2: Convert UTC to ST Algeria using the table (it states that we should add 1H to UTC for standard time so 0800ST Algeria)
► Refer to the table: An aircraft takes off from Guam at 2300 Standard Time on 30 April local date. After a flight of 11 HR 15 MIN it lands at Los Angeles (California). What is the standard time and local date of arrival (assume summer time rules apply)?
Answer 1715 on 30 April
1: Guam 23:00 ST 30/04 = 13:00 UTC 30/04 (using the table)
2: Add the flight time to this to get landing time in LA at 24:15 UTC 30/04
3: In the table it states that LA is UTC-8 = 16:15 ST 30/04
4: Don't forget that summer time rules apply!! (+1 H) = 17:15 ST 30/04
► On the 27th of February, at 52°S and 040°E, the sunrise is at 0243 UTC. On the same day, at 52°S and 035°W, the sunrise is at:
Answer: 0743 UTC
Difference in longitude = 75°
75° : 15°/H = 5H
Position 2 is to the West so lags behind = 0243 + 0500 = 0743 UTC
(Position 2 will get the sunrise on a later time than position 1)
► On 4 February the Air Almanac lists 1941 as the time of sunset at 5000S. An observer registers sunset at 2113 UTC this day. What is the observers position?
Answer: 5000S 02300W
Difference in time = 01:32
01:32 x 15°/H = 23°
The observer will have a later sunset so position is to the West (23°W)
► For 1 February the Air almanac lists the following data: Latitude: 6600N Morning civil twilight: 0756 Sunrise: 0900 Sunset: 1528 Evening civil twilight: 1632. The duration of morning twilight at 6600N is:
Answer: 1 hour 4 minutes and starts at 0756 LMT
Morning civil twilight (0756) to Sunrise (0900) = 01:04
► When the time is 20:00 UTC, it is:
Answer: 14:00 LMT at 90°West
90° : 15°/H = 6H
Position is West so lags behind UTC = 20:00 - 06:00 = 14:00
(This is the only option out of the 4 where it gives the correct result)
► What is the standard time in Hawaii when it is 06:00 ST on the 16th July in Queensland, Australia?
Answer: 10:00 ST 15 July
Queensland = ST - 10 to get UTC
Because we can not do 06:00 - 10:00 we add 24:00 and subtract 1 day:
30:00 ST 15/07 - 10:00 = 20:00 15/07 UTC
Hawaii ST = UTC-10
Hawaii ST = 10:00 15/07
► The local mean time at longitude 095°20minW, at 0000 UTC, is:
Answer: 17:38:40 Previous day
Position is 95°20' to the WEST of the prime meridian
(95°20') : 15°/H = 06:21:20 BEHIND UTC
24:00 (so that we can use it in calculations we added 24H and thus it will be -1 day)
24:00 - 06:21:20 = 17:38:40
► The SR/SS table for the 23rd of February at latitude 40°N gives:
SR = 06:44
SS = 17:44
At 12:00 Central European Time (UTC+1) at 40°N:
Answer: The sun rises at 64°W
12:00 CET = 11:00 UTC
Difference between 11:00UTC and 06:44LMT = 04:16
04:16 x 15°/H = 64°
► Mua, Tonga Islands, is situated at (21° 11minS, 175° 07minW). In the Air Almanac the standard time of Tonga Islands is listed as UTC + 13. For August 21st the sunrise table in the air almanac shows:
20°S: 06:18
30°S: 06:28
What is the standard time of sunrise at Mua
Answer: 06:59 on August 22nd
Inter polar between 20°S and 30°S to find the time of sunrise for 21°21'S = 06:19 LMT
The longitude is 175°07'W so (: 15°/h) = 11:40
Position is to the West so sunrise will occur LATER = 06:19 + 11:40 = 17:59UTC
Mua ST is listed as UTC+13
Mua ST = 17:59 + 13:00 = 30:59 ST 21/08
This is of course not possible so remove 24:00 and add 1 day = 06:59 ST 22/08
► An aircraft follows a radial to a VOR/DME station.
At 10:00 the DME reads 120 NM.
At 10:03 the DME reads 105 NM.
The estimated time overhead the VOR/DME station is:
Answer: 10:24:00
First find the groundspeed that the aircraft has (we covered 15NM in 3 minutes)
15 NM : 3 min = 300kt
Now use your groundspeed and the total distance to find the total time it takes
120 NM : 300kt = 24 min
Now add the flight time to your start time
10:00 + 00:24 = 10:24
► An aircraft departs from Schiphol airport and flies to Santa Cruz in Bolivia (South America) via Miami in Florida. The departure time (off blocks) is 07:45 ST at the 10th of november. Taxi time before take off at Schiphol is 25 minutes. The flight time to Miami Over the Atlantic Ocean is 09:20. The total taxi time in Miami to and from the gate is 25 minutes. The time spend at the gate is 02:40. From Miami to Santa Cruz the airborne time is 06:30. Calculate the time and date of touch down in Santa Cruz in ST Bolivia if the difference between ST and UTC is 5 hours.
Answer: 21:05 10th Nov.
Schiphol off blocks time = 07:45 ST , convert this to UTC (Schiphol is UTC+1)
Schiphol off blocks time = 06:45 UTC
(the question does not state the difference between UTC and ST as being 1h so remember it for the exams should it come up!
)Now you simply add up all the times in the order you read them in;
Off block = 06:45 UTC
+ 00:25 Taxi
= 07:10 UTC (take off time Schiphol)
+ 09:20 Flight time to Miami
= 16:30 UTC (landing time Miami)
+ 00:25 Taxi (both to and from Gate)
= 16:55 UTC
+ 02:40 Spend at Gate
= 19:35 UTC (take off time Miami)
+ 06:30 Flight time to Santa Cruz
= 26:05 UTC (landing in Santa Cruz)
Now convert to ST (given as "UTC-5") = 21:05 ST 10 Nov
