Anyone here knows how to do this??
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Anyone here knows how to do this??
An aircraft heading 017°(T) has a small island showing on the AWR at 45NM range on the 60° left azimuth line. To obtain a fix from this information you should plot:
A:
Range 45NM and QTE 060 from the center of the island
B:
Range 45NM and QTE 240 from the center of the island
C:
Range 45NM and QTE 317 from the center of the island
D:
Range 45NM and QTE 137 from the center of the island
AWR: I Assume it means Air Weather Radar....
Looking forward to hearing some ideas.
CG
A:
Range 45NM and QTE 060 from the center of the island
B:
Range 45NM and QTE 240 from the center of the island
C:
Range 45NM and QTE 317 from the center of the island
D:
Range 45NM and QTE 137 from the center of the island
AWR: I Assume it means Air Weather Radar....
Looking forward to hearing some ideas.
CG
How have you worked it out and what was your answer??
Well, you are heading 017 degrees T
The weather radar says the island is where?
whats QTE mean then? (QTE is the true bearing).
So??
Do a little sum.
Well, you are heading 017 degrees T
The weather radar says the island is where?
whats QTE mean then? (QTE is the true bearing).
So??
Do a little sum.
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whats QTE mean then? (QTE is the true bearing).
60° left azimuth line means, -60° relative bearing. so 017°T-060° = ?. As QTE is like a radial then answer is the reciprocal of that answer.
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Thank you for the answers.
I worked it out last night as well, and the correct answer is actually 137 as you need to plot a fix from the island per what I understood from the question. Q137 is what they say is the right answer.
I worked it out last night as well, and the correct answer is actually 137 as you need to plot a fix from the island per what I understood from the question. Q137 is what they say is the right answer.
Would love to know how Hakeem got B!!! 
I'd actually given the answer to the question already. I was trying to be helpful.
Funnily enough I got answer D.
whats QTE mean then? (QTE is the true bearing)
Funnily enough I got answer D.
Last edited by helimutt; 5th Apr 2010 at 15:04.
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helimutt, sorry to be a pedant, but you never actually gave the answer. You asked a series of leading questions but never actually gave any info, but I'm sure you knew what you meant.
Spendid Cruiser is correct and if you follow his workings throught you would get answer d.
Think of it in basic terms.
You know where the island is in relation to the aircraft heading (60 deg left @ 45Nm),
you know heading which is 017T,
therefore the bearing of the island from the aircraft is (017 degT -60deg=)317 degT @ 45Nm,
so you can therefore work out where the aircraft is in relation to the island (the reciprocal) 317degT - 180= 137 degT @ 45 Nm)
Once you have the reciprocal, you can plot that from the known position of the island (on a chart) to give the aircraft position.
It might help you if you draw a diagram North up, starting with the aircraft in the middle, and build it up as you work it out.
Hope this helps.
StraightLevel.
Spendid Cruiser is correct and if you follow his workings throught you would get answer d.
Think of it in basic terms.
You know where the island is in relation to the aircraft heading (60 deg left @ 45Nm),
you know heading which is 017T,
therefore the bearing of the island from the aircraft is (017 degT -60deg=)317 degT @ 45Nm,
so you can therefore work out where the aircraft is in relation to the island (the reciprocal) 317degT - 180= 137 degT @ 45 Nm)
Once you have the reciprocal, you can plot that from the known position of the island (on a chart) to give the aircraft position.
It might help you if you draw a diagram North up, starting with the aircraft in the middle, and build it up as you work it out.
Hope this helps.
StraightLevel.
I am sure that your specialist tutor at your groundschool will be able to:
a. confirm the answer and,
b. give you the reason why.
The question is from a very old CAA national ATPL examination.
If you ask such questions on this thread then God help you when you get the answers back.
LM is an ATPL consultant lecturer
a. confirm the answer and,
b. give you the reason why.
The question is from a very old CAA national ATPL examination.
If you ask such questions on this thread then God help you when you get the answers back.
LM is an ATPL consultant lecturer
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I also arrived at D , 017 - 60 and then - 180 (for the reciprocal) = 137
My old Rad Nav instructor was an ex RAF Tornado back seater and he explained all this very well i thought , although it does take hours of practice (i found anyway)
Just a note but there are lots of ADF and VOR tracking programmes on the net , visualising what the words are saying is very helpful , just google it.
My old Rad Nav instructor was an ex RAF Tornado back seater and he explained all this very well i thought , although it does take hours of practice (i found anyway)
Just a note but there are lots of ADF and VOR tracking programmes on the net , visualising what the words are saying is very helpful , just google it.
