Small Nav Q
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Small Nav Q
I would be most grateful for some advise and guidance as to the solution of the following Q which I have much struggled with but progressed none.
Thank you in advance for any help.
Ground feature observed on a relative bearing if 325 degs and five mins later on 280 degs.
A/C heading 165 (M) VAR 25 W DRIFT 10 right, and G/S 360 kts. When the relative brg was 280 degs, the distance and true brg of the A/C from the feature was?
Cron
Thank you in advance for any help.
Ground feature observed on a relative bearing if 325 degs and five mins later on 280 degs.
A/C heading 165 (M) VAR 25 W DRIFT 10 right, and G/S 360 kts. When the relative brg was 280 degs, the distance and true brg of the A/C from the feature was?
Cron
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Best way is to sketch a map of what is going on! However the bearing is very straight-forward, and I think will answer the question without you ever knowing the distance (there are 4 different bearings given if feedback from our students is accurate, although only 2 different distances).
Add relative bearing to magnetic heading to find magnetic bearing (since it comes to more than 360° you must subtract 360°). The answer is 085°M. Apply variation you find that this is 060°T, however that is from the aircraft to the feature, so you need the reciprocal which is 240°T, so you have half the answer.
On these questions, check the bearing change. If it is 45° as in this case you can bet the whole shape on your diagram will be an isosceles, right-angle triangle (the examiners love these). In this case that is true, because with the first bearing 35° left of the nose but with 10° stbd drift the bearing is 45° left of track. The second bearing is 80° left of the nose, or 90° left of the track.
You have an isosceles triangle so the distance from the second point to the feature is the same as the distance covered between the bearings (confirm this by looking at your diagram, that those are the two similar sides). The speed is 360 kts (6 nm per minute), time is 5 minutes so the aircraft travelled 30 nm. Therefore the distance from the feature is also 30 nm.
Distance 30 nm, bearing 240°T.
Note that if you find the distance, even with only a rough diagram the bearing is obvious, as the only other choice among the answers I believe are offered to go with 30 nm is the reciprocal bearing, 060°T.
Send Clowns
Gen Nav instructor, BCFT
Add relative bearing to magnetic heading to find magnetic bearing (since it comes to more than 360° you must subtract 360°). The answer is 085°M. Apply variation you find that this is 060°T, however that is from the aircraft to the feature, so you need the reciprocal which is 240°T, so you have half the answer.
On these questions, check the bearing change. If it is 45° as in this case you can bet the whole shape on your diagram will be an isosceles, right-angle triangle (the examiners love these). In this case that is true, because with the first bearing 35° left of the nose but with 10° stbd drift the bearing is 45° left of track. The second bearing is 80° left of the nose, or 90° left of the track.
You have an isosceles triangle so the distance from the second point to the feature is the same as the distance covered between the bearings (confirm this by looking at your diagram, that those are the two similar sides). The speed is 360 kts (6 nm per minute), time is 5 minutes so the aircraft travelled 30 nm. Therefore the distance from the feature is also 30 nm.
Distance 30 nm, bearing 240°T.
Note that if you find the distance, even with only a rough diagram the bearing is obvious, as the only other choice among the answers I believe are offered to go with 30 nm is the reciprocal bearing, 060°T.
Send Clowns
Gen Nav instructor, BCFT
