CNS Exercises with answers
Week 1
Lateral Position
1. What is the EHAM/AMS WGS84 Lat/Long coordinates in 3 formats:
Amsterdam AMS (Basic) = 52°18’29”N/ 004°45’51”E
Amsterdam AMS (FMS) = 52°18,5’ N/ 004°45,8’ E
From Minutes/Seconds to decimal Minutes:
18+29/60 = 18,48333
45+51/60 = 45,85
Amsterdam AMS (Mathematica) =52,3081°N004,7642°E
1° = 60 minutes
From Minutes to decimal Degrees:
52 + 18, = 52 + + = 52,3081
4 + 45, = 4 + + = 4,7642
2. Prove that 1 nm = 1852m, use the circumference of the earth ± 40000 km by definition, a Nautical
mile = distance of 1 minute great circle (GC)
Circumference of the Earth is (about) 40000km – for 360°
1 NM is 1 minute of angle
And 1 minute of angle is 1/60 of 1°
Thus, 1 NM = 40000*1000/(360*60) = 1852 m
3. What is the accuracy of the position in format 1 (normal WGS84 format)?
WGS84 format: 1 second of 1°
1NM for 1 minute
So, for 1 second: 1852/60 = 31m
4. What is the accuracy of the position in format 2 (FMC format)?
FMC format: 1 decimal (1/10) of a minute
1NM for 1 minute
So, for 1/10 of a minute: 1852/10 = 185m
5. What is the accuracy of the position in format 3 (Engineering format)?
Engineering format: 4 decimals (1/10000) of 1 degree
1NM for 1 minute 60NM for 1 degree
So, for 1/10000 of a degree: 1852*60/10000 = 11m
GC Distance
6. Find the shortest GC track and distance in nm from position 00⁰N 168⁰W to 00⁰N 173⁰E:
Along equator 19°
There are 60 minutes in 1 degree, and 1NM per minute, so 1 degree corresponds to 60NM.
So, 19 * 60NM = 1140NM
7. Find the shortest GC track and distance in nm from position 60⁰N 168⁰W to 60⁰N 173⁰E:
Change in Latitude of 60°N, so, distance is reduced by Cos(60°) = 0.5
19° * 0.5 = 9.5°, 9.5 * 60NM = 570NM
8. Find the shortest GC track and distance in nm from position 60⁰N 168⁰E to 60⁰S 168⁰E:
Along Longitude for 120°, so, 120 * 60NM = 7200NM
Week 1
Lateral Position
1. What is the EHAM/AMS WGS84 Lat/Long coordinates in 3 formats:
Amsterdam AMS (Basic) = 52°18’29”N/ 004°45’51”E
Amsterdam AMS (FMS) = 52°18,5’ N/ 004°45,8’ E
From Minutes/Seconds to decimal Minutes:
18+29/60 = 18,48333
45+51/60 = 45,85
Amsterdam AMS (Mathematica) =52,3081°N004,7642°E
1° = 60 minutes
From Minutes to decimal Degrees:
52 + 18, = 52 + + = 52,3081
4 + 45, = 4 + + = 4,7642
2. Prove that 1 nm = 1852m, use the circumference of the earth ± 40000 km by definition, a Nautical
mile = distance of 1 minute great circle (GC)
Circumference of the Earth is (about) 40000km – for 360°
1 NM is 1 minute of angle
And 1 minute of angle is 1/60 of 1°
Thus, 1 NM = 40000*1000/(360*60) = 1852 m
3. What is the accuracy of the position in format 1 (normal WGS84 format)?
WGS84 format: 1 second of 1°
1NM for 1 minute
So, for 1 second: 1852/60 = 31m
4. What is the accuracy of the position in format 2 (FMC format)?
FMC format: 1 decimal (1/10) of a minute
1NM for 1 minute
So, for 1/10 of a minute: 1852/10 = 185m
5. What is the accuracy of the position in format 3 (Engineering format)?
Engineering format: 4 decimals (1/10000) of 1 degree
1NM for 1 minute 60NM for 1 degree
So, for 1/10000 of a degree: 1852*60/10000 = 11m
GC Distance
6. Find the shortest GC track and distance in nm from position 00⁰N 168⁰W to 00⁰N 173⁰E:
Along equator 19°
There are 60 minutes in 1 degree, and 1NM per minute, so 1 degree corresponds to 60NM.
So, 19 * 60NM = 1140NM
7. Find the shortest GC track and distance in nm from position 60⁰N 168⁰W to 60⁰N 173⁰E:
Change in Latitude of 60°N, so, distance is reduced by Cos(60°) = 0.5
19° * 0.5 = 9.5°, 9.5 * 60NM = 570NM
8. Find the shortest GC track and distance in nm from position 60⁰N 168⁰E to 60⁰S 168⁰E:
Along Longitude for 120°, so, 120 * 60NM = 7200NM
