Navigation Questions
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Navigation Questions
So I can't figure out few Q's about variation and deviation.
I need to know this by tommorow.
CH is 120°. Deviation is +4°. Variation is 5° W. Wind correction angle is 8° Right. TT is: (choose below)
111°
127°
113°
129°
CH is 224°. Deviation is -3°. Variation is 4° W. TH is:
217°
223°
225°
231°
CH is 317°. Deviation is +3°. Variation is 5° W. Wind correction angle is 10° right. TT is:
309°
305°
329°
325°
CH is 098°. Deviation is +2°. Variation is -8° and drift is 5° Left. TT is:
087°
093°
097°
109°
So what is the rule?
It might be easy for you guys but I'm stuck with these.
I still hav few questions left but I need to search for it.
Thanks for helping out!
Distances
Calculate the distance AB via the shortest route. A: 160°56'E and 089°54'N, B: 019°04'W and 066°13'N.
2 points are located on the same longitude and on 34° 30' N and 26° 20' S respectively. The distance between both points is:
I need to know this by tommorow.
CH is 120°. Deviation is +4°. Variation is 5° W. Wind correction angle is 8° Right. TT is: (choose below)
111°
127°
113°
129°
CH is 224°. Deviation is -3°. Variation is 4° W. TH is:
217°
223°
225°
231°
CH is 317°. Deviation is +3°. Variation is 5° W. Wind correction angle is 10° right. TT is:
309°
305°
329°
325°
CH is 098°. Deviation is +2°. Variation is -8° and drift is 5° Left. TT is:
087°
093°
097°
109°
So what is the rule?
It might be easy for you guys but I'm stuck with these.
I still hav few questions left but I need to search for it.
Thanks for helping out!
Distances
Calculate the distance AB via the shortest route. A: 160°56'E and 089°54'N, B: 019°04'W and 066°13'N.
2 points are located on the same longitude and on 34° 30' N and 26° 20' S respectively. The distance between both points is:
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Look outside and steer towards that big city / lake / whatever you use to navigate. A track error of 2 degrees is bullsh*t in a small airplane if there is any variable wind.
Or use GPS...
Or use GPS...
Most questions have been asked on pprune before, so using the search function will often get you the information that you need. The link below may enable you to solve your compass problems.
http://www.pprune.org/professional-p...+TRUE+headings
http://www.pprune.org/professional-p...+TRUE+headings
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Pirke
With a reply like that to a simple request for help from an aspiring aviator I am tempted to tell you to Get Lost.
However with your attitude to navigation you have demonstrated you are quite capable of getting lost without assistance from anyone else.
Syriadh
Try this for a little light relief. ......... Most of the advice is as good today as when the film was made
http://youtu.be/C6oGa1bqe1U
However with your attitude to navigation you have demonstrated you are quite capable of getting lost without assistance from anyone else.
Syriadh
Try this for a little light relief. ......... Most of the advice is as good today as when the film was made
http://youtu.be/C6oGa1bqe1U
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CH is 120°. Deviation is +4°. Variation is 5° W. Wind correction angle is 8° Right. TT is: (choose below)
111°
127°
113°
129°
120-4-5-8=103 (T) ???
CH is 224°. Deviation is -3°. Variation is 4° W. TH is:
217°
223°
225°
231°
224+3-4 = 223(T) = answer b
CH is 317°. Deviation is +3°. Variation is 5° W. Wind correction angle is 10° right. TT is:
309°
305°
329°
325°
317-3-5-10 = 299 (T) ???
CH is 098°. Deviation is +2°. Variation is -8° and drift is 5° Left. TT is:
087°
093°
097°
109°
098-2+8+5 = 109 (T) = answer d
have I gone wrong somewhere on questions 1 and 3 ???
111°
127°
113°
129°
120-4-5-8=103 (T) ???
CH is 224°. Deviation is -3°. Variation is 4° W. TH is:
217°
223°
225°
231°
224+3-4 = 223(T) = answer b
CH is 317°. Deviation is +3°. Variation is 5° W. Wind correction angle is 10° right. TT is:
309°
305°
329°
325°
317-3-5-10 = 299 (T) ???
CH is 098°. Deviation is +2°. Variation is -8° and drift is 5° Left. TT is:
087°
093°
097°
109°
098-2+8+5 = 109 (T) = answer d
have I gone wrong somewhere on questions 1 and 3 ???
Backwards?
I'm intrigued by these questions. I've been in some form of adult technical/nuclear education most of my life. In my experience test questions come from objectives that come from job/task analysis - "what does the individual have to know to do the job?"
A training objective might be: "Given a true course, an aeronautical chart, and a compass deviation card, CALCULATE the required compass heading required for a flight."
In the real world you draw a pencil line on the chart, measure the angle in degrees (true) then apply the magic (TVMDC+W) to find what number you want to look at in the cockpit.
These questions all look like they start with the compass heading and ask you to find the true course. If so, this is backwards from what actually happens. This is not a skill required of a pilot.
Am I missing something?
Terry
A training objective might be: "Given a true course, an aeronautical chart, and a compass deviation card, CALCULATE the required compass heading required for a flight."
In the real world you draw a pencil line on the chart, measure the angle in degrees (true) then apply the magic (TVMDC+W) to find what number you want to look at in the cockpit.
These questions all look like they start with the compass heading and ask you to find the true course. If so, this is backwards from what actually happens. This is not a skill required of a pilot.
Am I missing something?
Terry
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In the real world you draw a pencil line on the chart, measure the angle in degrees (true) then apply the magic (TVMDC+W) to find what number you want to look at in the cockpit.
have I gone wrong somewhere on questions 1 and 3 ???
As far as I can see, only Question 2. has the correct answer in the alternatives.
Distances
Calculate the distance AB via the shortest route. A: 160°56'E and 089°54'N, B: 019°04'W and 066°13'N.
Calculate the distance AB via the shortest route. A: 160°56'E and 089°54'N, B: 019°04'W and 066°13'N.
2 points are located on the same longitude and on 34° 30' N and 26° 20' S respectively. The distance between both points is:
A track error of 2 degrees is bullsh*t in a small airplane if there is any variable wind.
"who cares?".
Am I missing something? Terry
MJ
Last edited by Mach Jump; 30th Dec 2014 at 00:03. Reason: Added 'Distances'
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Mach Jump
Thank you for the answers. Very strange that the correct answer is not given (except for no 2).
I concur that the normal procedure would be from Chart (T) to Compass, however its an exam question so it's testing the reader's knowledge. Though I think a bit unfair not to offer the correct answer, except no 2 (I checked and re-checked my calculations several times).
I've not seen a drift angle given before in questions, normally its an off-track distance after cruising 30/60/90/120 nm, where the 1 in 60 rule is used to calculate a correction heading.
all good fun for the classroom .
flyme
Thank you for the answers. Very strange that the correct answer is not given (except for no 2).
I concur that the normal procedure would be from Chart (T) to Compass, however its an exam question so it's testing the reader's knowledge. Though I think a bit unfair not to offer the correct answer, except no 2 (I checked and re-checked my calculations several times).
I've not seen a drift angle given before in questions, normally its an off-track distance after cruising 30/60/90/120 nm, where the 1 in 60 rule is used to calculate a correction heading.
all good fun for the classroom .
flyme
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CH is 120°. Deviation is +4°. Variation is 5° W. Wind correction angle is 8° Right. TT is: (choose below)
111°
127°
113°
129°
Err... isnt the correct answer 111?
CH = 120, Deviation is +4 (E), so MH = 124 as Compass North is 4 east of Mag North.
Variation = 5W so TH = 119 as Mag North is 5 west of TN.
That leaves WCA of 8 right, so you are flying an extra 8 degrees right to compensate for the wind, giving a TT of 119 -8?
Youd need to recalc the others!
111°
127°
113°
129°
Err... isnt the correct answer 111?
CH = 120, Deviation is +4 (E), so MH = 124 as Compass North is 4 east of Mag North.
Variation = 5W so TH = 119 as Mag North is 5 west of TN.
That leaves WCA of 8 right, so you are flying an extra 8 degrees right to compensate for the wind, giving a TT of 119 -8?
Youd need to recalc the others!
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CH is 120°. Deviation is +4°. Variation is 5° W. Wind correction angle is 8° Right. TT is: (choose below)
111°
127°
113°
129°
120+4-5-8=111 (T) answer a
CH is 224°. Deviation is -3°. Variation is 4° W. TH is:
217°
223°
225°
231°
224-3-4 = 217 (T) = answer a
CH is 317°. Deviation is +3°. Variation is 5° W. Wind correction angle is 10° right. TT is:
309°
305°
329°
325°
317-3-5-10 = 305 (T) answer b
CH is 098°. Deviation is +2°. Variation is -8° and drift is 5° Left. TT is:
087°
093°
097°
109°
098-2+8+5 = 087 (T) = answer a
the only problem is I would normally have applied the deviation in the opposite sense . . . .
----------------
got all my books out and rechecked answers: setting it out in the normal way:-
103 T +4+5+8 = 120 C
223 T -3+4 = 224 C
299 T +3+5+10 = 317 C
109 T +2-8-5 = 098 C
only question 2 provides the correct answer.
111°
127°
113°
129°
120+4-5-8=111 (T) answer a
CH is 224°. Deviation is -3°. Variation is 4° W. TH is:
217°
223°
225°
231°
224-3-4 = 217 (T) = answer a
CH is 317°. Deviation is +3°. Variation is 5° W. Wind correction angle is 10° right. TT is:
309°
305°
329°
325°
317-3-5-10 = 305 (T) answer b
CH is 098°. Deviation is +2°. Variation is -8° and drift is 5° Left. TT is:
087°
093°
097°
109°
098-2+8+5 = 087 (T) = answer a
the only problem is I would normally have applied the deviation in the opposite sense . . . .
----------------
got all my books out and rechecked answers: setting it out in the normal way:-
103 T +4+5+8 = 120 C
223 T -3+4 = 224 C
299 T +3+5+10 = 317 C
109 T +2-8-5 = 098 C
only question 2 provides the correct answer.
Last edited by flyme273; 31st Dec 2014 at 07:48. Reason: correction
I've been in some form of adult technical/nuclear education most of my life. In my experience test questions come from objectives that come from job/task analysis - "what does the individual have to know to do the job?"
The JAA produced a central question bank for commercial exams and then from the questions, someone else formulated objectives! All backwards of course.
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got all my books out and rechecked answers: setting it out in the normal way:-
103 T +4+5+8 = 120 C
223 T -3+4 = 224 C
299 T +3+5+10 = 317 C
109 T +2-8-5 = 098 C
only question 2 provides the correct answer.
103 T +4+5+8 = 120 C
223 T -3+4 = 224 C
299 T +3+5+10 = 317 C
109 T +2-8-5 = 098 C
only question 2 provides the correct answer.
Westerly values can be written as 5W or -5 degrees. Easterly are 5E or +5 degrees. Convert all the numbers your given to E/W format (+4 = 4E) and apply the rule of thumb.
So working forward gives
Q1 = 111 + 8 + 5(West is best) - 4(East is least) = 120
I've no idea where you got the questions from, but how likely is it that 3 out of 4 questions don't give you the correct option...?
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Tractor Boy
Westerly values can be written as 5W or -5 degrees. Easterly are 5E or +5 degrees.
sorry do not agree. Should be West is (best) positive +; East is (least) negative.
flyme.
Westerly values can be written as 5W or -5 degrees. Easterly are 5E or +5 degrees.
sorry do not agree. Should be West is (best) positive +; East is (least) negative.
flyme.
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Flyme273 - I can assure you that Westerly variations are written -ve, Easterly are +ve
One of many worked examples...
HDG Question - ATP Forum
Edit: this one explains it a bit more
http://www.atpforum.eu/showthread.php?t=8850
I went through all this myself doing GNav on the ATPL course
One of many worked examples...
HDG Question - ATP Forum
Edit: this one explains it a bit more
http://www.atpforum.eu/showthread.php?t=8850
I went through all this myself doing GNav on the ATPL course
Last edited by TractorBoy; 3rd Jan 2015 at 11:46.
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Hello there!
It's true, we live in a GPS age, so all those question about TT WCA CH etc. have more practical relevance before the examination table than when flying.
What we usually know from the map is TT from A to B, and what we want to know is CH.
There is a single fomula that takes care of everything. It goes like this:
CH = TT PLUS WCA MINUS VARIATION MINUS DEVIATION.
Do not forget: wca as well as var. and dev. can all have positive or negative values, if the crosswind blows from the left, the wca is negative, and if magnetic north lies left (west) of true north the var. is negative etc...
An example: TT=265, WCA= -20, VAR west 3, DEV east 4.
CH=265 +(-20)-(-3)-(+4)
=265-20+3-4
=244.
It's true, we live in a GPS age, so all those question about TT WCA CH etc. have more practical relevance before the examination table than when flying.
What we usually know from the map is TT from A to B, and what we want to know is CH.
There is a single fomula that takes care of everything. It goes like this:
CH = TT PLUS WCA MINUS VARIATION MINUS DEVIATION.
Do not forget: wca as well as var. and dev. can all have positive or negative values, if the crosswind blows from the left, the wca is negative, and if magnetic north lies left (west) of true north the var. is negative etc...
An example: TT=265, WCA= -20, VAR west 3, DEV east 4.
CH=265 +(-20)-(-3)-(+4)
=265-20+3-4
=244.
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From a very deep portion of my fading memory, I come up with the following:
C D M V T
120 +4 124 -5 119 +8 127
224 -3 221 -4 217
317 +3 320 -5 315 -10 325
098 +2 100 - 8 097 -5 087
These are one of those 'plug in the numbers' types types of tests.
I believe MachJump has the distance questions hacked. As ever, 1' equals one mile and a diagram helps. Also remember, in real life and short distances 500 nm or so, Pythagoras works a treat.
PM
C D M V T
120 +4 124 -5 119 +8 127
224 -3 221 -4 217
317 +3 320 -5 315 -10 325
098 +2 100 - 8 097 -5 087
These are one of those 'plug in the numbers' types types of tests.
I believe MachJump has the distance questions hacked. As ever, 1' equals one mile and a diagram helps. Also remember, in real life and short distances 500 nm or so, Pythagoras works a treat.
PM