# Navigation Questions

Thread Starter

Join Date: Nov 2013

Location: Antwerp

Age: 22

Posts: 6

**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:

Join Date: Jan 2011

Location: England

Posts: 643

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

Join Date: Jan 1999

Location: north of barlu

Posts: 6,179

**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.

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

Join Date: Jul 2007

Location: Netherlands

Posts: 122

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 ???

Join Date: Dec 2011

Location: Augusta, Georgia, USA (back from Germany again)

Posts: 147

**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

Join Date: Nov 2000

Location: Cambridge, England, EU

Posts: 3,437

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.

Join Date: May 2005

Location: Yorkshire

Posts: 1,085

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 01:03. Reason: Added 'Distances'*

Join Date: Jul 2007

Location: Netherlands

Posts: 122

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

Join Date: Aug 2005

Location: EGSX

Age: 52

Posts: 177

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!

Join Date: Jul 2007

Location: Netherlands

Posts: 122

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 08:48. Reason: correction*

Join Date: Oct 2004

Location: Europe

Posts: 5,973

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.

Join Date: Jun 2001

Location: Wherever someone will pay me to do fun stuff

Posts: 1,124

Sorry in advance for the wild thread drift.

Sadly it happens all too often, and probably equally often means that something that people need to know is overlooked because there's no question on the topic.

Add to the mix, that by the time people get involved is setting questions and the like it's very easy to forget that so many things that are now taken for granted had to be learned at some point, and the unending desire to accelerate training and it's no wonder that after a mishap or accident the basic competence of the crew is so commonly questioned these days!

The JAA produced a central question bank for commercial exams and then from the questions, someone else formulated objectives!

Add to the mix, that by the time people get involved is setting questions and the like it's very easy to forget that so many things that are now taken for granted had to be learned at some point, and the unending desire to accelerate training and it's no wonder that after a mishap or accident the basic competence of the crew is so commonly questioned these days!

Join Date: Aug 2005

Location: EGSX

Age: 52

Posts: 177

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...?

Join Date: Aug 2005

Location: EGSX

Age: 52

Posts: 177

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 12:46.*

Join Date: Mar 2015

Location: denmark

Posts: 1

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.