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► An NDB is located at position (N55°26min, W005°42min). The variation at the NDB is 9°W. The position of the aircraft is (56°00minN, 010°00minW). The variation at the aircraft position is 11°W. The initial TT of the great circle from the aircraft position to the NDB position is 101,5°. What is the Magnetic Bearing of the NDB from the aircraft?
Answer: 112,5°
True track TO NDB = 101.5°
Magnetic bearing TO NDB = (101.5° + 11°VAR) 112.5°
► A VOR is situated at position (N55°26min, W005°42min). The variation at the VOR is 9°W. The position of the aircraft is (N60°00min, W010°00min). The variation at the aircraft position is 11°W. The initial TT angle of the great circle from the aircraft position to the VOR is 101.5°. Which radial is the aircraft on?
Answer: 294°
Difference between true track at the aircraft and true track at VOR is the convergency
Convergency = Ch Long (10° - 5°42') x Sin((60° + 55°26') / 2)
Convergency = 3.6°
If you draw the sketch with the 2 meridians in the Northern Hemisphere we can see that the true track at the VOR is larger than the true track at the aircraft, thus:
TT Aircraft (101.5°) + Convergency (3.6°) = TT at VOR (105°)
They ask us the radial which is FROM the VOR so take the reciprocal of 105.1 = 285°
For a VOR we take the variation AT the VOR (for a NDB we take it at the aircraft) so 285° + 9°W = 294°
► Given: Position NDB (55°10minN, 012°55minE)
DR Position (54°53minN, 009°58minE)
NDB on the RMI reads 090°
Magnetic variation 10°W
The position line has to be plotted on a Lamberts Conformal Chart with standard parallels at 40°N and 48°N. Calculate the direction (T) of the bearing to be plotted from the NDB:
Answer: 262°
An RMI reads FROM the aircraft TO the NDB
1 - Apply Variation
For a NDB we take the variation at our position (10°W). Apply that to the Magnetic 90° to get a TRUE track of 80°(T).
2 - Consider convergency (to plot on the chart)
Constant of cone = Sin(parallel of origin)
Constant of cone = Sin(40+48 /2)
Constant of cone = 0.69
Convergency = Ch long (2°57') x 0.69
Convergency = 2
If we draw the sketch of the 2 meridians and our track we can see that the True Track at the NDB station is bigger so 80°(T) + 2° = 82°(T) at the NDB
The last thing we need to do is turn it around because they ask us the bearing FROM the NDB so 82 + 180 = 262°(T)
► An aircraft is over position HO (55°30minN 060°15minW), where YYR VOR (53°30minN 060°15minW) can be received. The magnetic variation is 31°W at HO and 28°W at YYR. What is the radial from YYR?
Answer: 028°
Both are on the same longitudes, the VOR is to the South (180°) of the aircraft thus the aircraft is on true bearing North from the VOR. For VOR's we take the variation at the station so 000° + 28°W = 028°
► Given:
CON VOR (53°54.8minN 008°49.1minW) DME 30NM
CRN DME (53°18.1minN 008°56.5minW) DME 25NM
Aircraft heading is 270°(M) and both DME distances are decreasing. What is the aircraft position?
Answer: 53°30minN 008°20minW
Take a pair of dividers and open them to a distance of 30NM (measure against latitude scale on the chart). Now draw a circle around the CON VOR. Repeat these steps for the CRN DME (but now with the 25NM).
We can now see that both circles intersect at 2 points (1 east and 1 west of the stations). The question states that distances are decreasing (we are flying towards the stations with heading 270°) so we can only be at the intersection on the East side.
► Given true heading 256°, VAR 13°E, relative bearing to a station is 333°. The true bearing to the station is:
Answer: 229°
TH (256°) + RB (333°) = 229° TB (589-360)
Answer: 112,5°
True track TO NDB = 101.5°
Magnetic bearing TO NDB = (101.5° + 11°VAR) 112.5°
► A VOR is situated at position (N55°26min, W005°42min). The variation at the VOR is 9°W. The position of the aircraft is (N60°00min, W010°00min). The variation at the aircraft position is 11°W. The initial TT angle of the great circle from the aircraft position to the VOR is 101.5°. Which radial is the aircraft on?
Answer: 294°
Difference between true track at the aircraft and true track at VOR is the convergency
Convergency = Ch Long (10° - 5°42') x Sin((60° + 55°26') / 2)
Convergency = 3.6°
If you draw the sketch with the 2 meridians in the Northern Hemisphere we can see that the true track at the VOR is larger than the true track at the aircraft, thus:
TT Aircraft (101.5°) + Convergency (3.6°) = TT at VOR (105°)
They ask us the radial which is FROM the VOR so take the reciprocal of 105.1 = 285°
For a VOR we take the variation AT the VOR (for a NDB we take it at the aircraft) so 285° + 9°W = 294°
► Given: Position NDB (55°10minN, 012°55minE)
DR Position (54°53minN, 009°58minE)
NDB on the RMI reads 090°
Magnetic variation 10°W
The position line has to be plotted on a Lamberts Conformal Chart with standard parallels at 40°N and 48°N. Calculate the direction (T) of the bearing to be plotted from the NDB:
Answer: 262°
An RMI reads FROM the aircraft TO the NDB
1 - Apply Variation
For a NDB we take the variation at our position (10°W). Apply that to the Magnetic 90° to get a TRUE track of 80°(T).
2 - Consider convergency (to plot on the chart)
Constant of cone = Sin(parallel of origin)
Constant of cone = Sin(40+48 /2)
Constant of cone = 0.69
Convergency = Ch long (2°57') x 0.69
Convergency = 2
If we draw the sketch of the 2 meridians and our track we can see that the True Track at the NDB station is bigger so 80°(T) + 2° = 82°(T) at the NDB
The last thing we need to do is turn it around because they ask us the bearing FROM the NDB so 82 + 180 = 262°(T)
► An aircraft is over position HO (55°30minN 060°15minW), where YYR VOR (53°30minN 060°15minW) can be received. The magnetic variation is 31°W at HO and 28°W at YYR. What is the radial from YYR?
Answer: 028°
Both are on the same longitudes, the VOR is to the South (180°) of the aircraft thus the aircraft is on true bearing North from the VOR. For VOR's we take the variation at the station so 000° + 28°W = 028°
► Given:
CON VOR (53°54.8minN 008°49.1minW) DME 30NM
CRN DME (53°18.1minN 008°56.5minW) DME 25NM
Aircraft heading is 270°(M) and both DME distances are decreasing. What is the aircraft position?
Answer: 53°30minN 008°20minW
Take a pair of dividers and open them to a distance of 30NM (measure against latitude scale on the chart). Now draw a circle around the CON VOR. Repeat these steps for the CRN DME (but now with the 25NM).
We can now see that both circles intersect at 2 points (1 east and 1 west of the stations). The question states that distances are decreasing (we are flying towards the stations with heading 270°) so we can only be at the intersection on the East side.
► Given true heading 256°, VAR 13°E, relative bearing to a station is 333°. The true bearing to the station is:
Answer: 229°
TH (256°) + RB (333°) = 229° TB (589-360)
