ATP question
Thread Starter
Join Date: Dec 2017
Location: Ho Chi Minh
Posts: 1
Likes: 0
Received 0 Likes
on
0 Posts
ATP question
Is someone able to help me to explain these two questions please? Many thanks!
1. An INS-equipped aircraft flies from 56N 020W (waypoint 3) to 56N 030W (waypoint 4). The initial INS desired track at waypoint 3 is:
A. 086 (T)
B. 082 (T)
C. 274 (T) (correct)
D. 278 (T)
2. An aircraft starts at position 0410S 17822W and tracks true north for 2950nm, then turn 99 degrees left, and maintains a rhumb line track for for 314 km. What is the final position?
A. 5500N 17422W
B. 4500N 17422W
C. 5509N 17738E
D. 4500N 17738E (correct)
1. An INS-equipped aircraft flies from 56N 020W (waypoint 3) to 56N 030W (waypoint 4). The initial INS desired track at waypoint 3 is:
A. 086 (T)
B. 082 (T)
C. 274 (T) (correct)
D. 278 (T)
2. An aircraft starts at position 0410S 17822W and tracks true north for 2950nm, then turn 99 degrees left, and maintains a rhumb line track for for 314 km. What is the final position?
A. 5500N 17422W
B. 4500N 17422W
C. 5509N 17738E
D. 4500N 17738E (correct)
Join Date: Jul 2003
Location: UK
Posts: 461
Likes: 0
Received 0 Likes
on
0 Posts
annampilot,
Q2 can be worked out using mental maths.
The aircraft is tracking due North, therefore following the 17822W meridian. It flies for 2950Nm which 1 degree being 60Nm equals approx 49 degrees, so from 4 degrees South it flies to 45 degrees north, now you have a choice of 2 answers. So it turns left 99 degrees (heading slightly south of further West) for 314Km which is approx 169Nm, at 45 degrees North it will cross the 180 meridian this now becomes an Eastern position, which can only be answer D. No need for plotting chart or completely accurate calculations just a pencil and paper to draw(if need be) the course of the aircraft. These sort of questions sound complex, but looking at the simplistic way of doing it can save you time during an exam, try not to look too deep into them.
Q2 can be worked out using mental maths.
The aircraft is tracking due North, therefore following the 17822W meridian. It flies for 2950Nm which 1 degree being 60Nm equals approx 49 degrees, so from 4 degrees South it flies to 45 degrees north, now you have a choice of 2 answers. So it turns left 99 degrees (heading slightly south of further West) for 314Km which is approx 169Nm, at 45 degrees North it will cross the 180 meridian this now becomes an Eastern position, which can only be answer D. No need for plotting chart or completely accurate calculations just a pencil and paper to draw(if need be) the course of the aircraft. These sort of questions sound complex, but looking at the simplistic way of doing it can save you time during an exam, try not to look too deep into them.
Join Date: Oct 2007
Location: Wherever I go, there I am
Age: 43
Posts: 804
Likes: 0
Received 0 Likes
on
0 Posts
A similar process of stepping back to look at the question can be used for question one too.
Ask what is common between the two coordinates?
56N 020W and 56N 030W.
You’ve flown the same latitude but gained 10 degrees longitude. You have to be flying west; and as close to due west as possible as you didn’t increase or decrease your latitude (i.e., go to 56.xN or 55.xN).
So, 274T would be “more correct” than 278T as it’s closer to due west. No doubt someone has some wicked cool math formula that will specifically tell you 274T is the correct track, and that can be used to back-up your mental image of the question and answer sets.
The filler about the INS is just that; filler. It’s meant to distract you from what is a basic navigation question asked at a PPL level.
To make sure you understand this, reverse the question. What would be the initial track if you start out from 55N 020W and flew to 55N 010W?
Now let’s throw a spanner in - what is the track if you fly from 56N 020W to 56N 020E. Give the answer for both the short and the long way around.
Let’s try something else. What is the track if you fly from 56N 020W to 55N 020W or to 56S 020W.
Do these and you’ll start to see that the question really is basic, and they could substitute FMC, FMS, IRS, INS, et cettera to make it sound “ATPL-y.”
Ask what is common between the two coordinates?
56N 020W and 56N 030W.
You’ve flown the same latitude but gained 10 degrees longitude. You have to be flying west; and as close to due west as possible as you didn’t increase or decrease your latitude (i.e., go to 56.xN or 55.xN).
So, 274T would be “more correct” than 278T as it’s closer to due west. No doubt someone has some wicked cool math formula that will specifically tell you 274T is the correct track, and that can be used to back-up your mental image of the question and answer sets.
The filler about the INS is just that; filler. It’s meant to distract you from what is a basic navigation question asked at a PPL level.
To make sure you understand this, reverse the question. What would be the initial track if you start out from 55N 020W and flew to 55N 010W?
Now let’s throw a spanner in - what is the track if you fly from 56N 020W to 56N 020E. Give the answer for both the short and the long way around.
Let’s try something else. What is the track if you fly from 56N 020W to 55N 020W or to 56S 020W.
Do these and you’ll start to see that the question really is basic, and they could substitute FMC, FMS, IRS, INS, et cettera to make it sound “ATPL-y.”
Last edited by +TSRA; 30th Dec 2017 at 17:33. Reason: Wordings
Join Date: Jun 2014
Location: Earth
Posts: 666
Likes: 0
Received 0 Likes
on
0 Posts
extremely basic questions, the most simple application of the conversion angle formula for the first and a knowledge of 60nm being equal to 1 degree of longitude should have the second solved for you ... cba to do it but it's not something that warrants a thread in all honesty
Join Date: Oct 2007
Location: Wherever I go, there I am
Age: 43
Posts: 804
Likes: 0
Received 0 Likes
on
0 Posts
Ease up a little there Officer Kite; whats the point of having a forum for Professional Pilot Studies if no one can ask a question, even those that are basic to those of us in industry for a while now? “Basic” is relative to experience and previous knowledge, so what is basic to you and I might me relatively advanced concepts for someone else, hence the beauty of being able to ask questions in an open environment like PPRuNe.
Join Date: Jun 2014
Location: Earth
Posts: 666
Likes: 0
Received 0 Likes
on
0 Posts
Apologies, i didn't mean to come across as dismissive or rude towards asking a question. I am an ATPL student myself and have used this forum in the past for difficult concepts. It's just that there are many things in Gnav I would vouch for being difficult to 'click', but the stuff in the first post is not difficult at all for anybody who has opened the book/cbt and looked at the chapter in question. They're the types of questions of which there are numerous worked examples too, it just felt like they hadn't bothered to try look at it and understand and just wanted a quick answer was all.
Join Date: Aug 2013
Location: Spain
Posts: 165
Likes: 0
Received 0 Likes
on
0 Posts
θ = atan2( sin Δλ ⋅ cos φ2 , cos φ1 ⋅ sin φ2 − sin φ1 ⋅ cos φ2 ⋅ cos Δλ )
where:
θ is the initial bearing,
φ1 is the start point,
φ2 the end point,
Δλ is the difference in longitude
θ = atan2( sin (-10) ⋅ cos (56.03) , cos (56.02) ⋅ sin (56.03) − sin (56.02) ⋅ cos (56.03) ⋅ cos (-10) )
θ = atan2(-0.09702743761, 0.00721372431)
θ = -1.4965855976348807 radians which is -85.7480384 degrees
or (-85.7480384 + 360) = 274.25 degrees
Join Date: Nov 2000
Location: White Waltham, Prestwick & Calgary
Age: 72
Posts: 4,149
Likes: 0
Received 29 Likes
on
14 Posts
Can't you just half the convergency of 8.29 degs? sin 56 (.829) x 10
All you have to do is remember which way to apply it - and rum comes from Jamaica
All you have to do is remember which way to apply it - and rum comes from Jamaica
Join Date: Jun 2014
Location: Earth
Posts: 666
Likes: 0
Received 0 Likes
on
0 Posts
Fair enough well OK here it goes.
Q 1 ->
For this we need to use the conversion angle formula.
CA = 1/2 (Change in Longitude x SIN Latitude)
From the question we can establish our longitude will change by 10 degrees and our latitude is 56N
So back to the formula ... CA = 1/2 (10 x SIN 56) ... = 1/2 (8.29) = 4.1 deg ... so 4 degrees.
Now we have to refer back to the question to decide what we're gonna do with this 4 degrees. Once you establish we are heading west along the same line of latitude we can say we are heading 270 degrees. We are also in the northern hemisphere and heading west, this means when departing we are gonna head further north than simply following the rhumb line of the 56N latitude line, this shortens our journey time, we know exactly how much we need to turn north of the 56N line of latitude by using the CA formula as above. So adding 270 + 4 = 274. Again, this will be our heading on departure and will be constantly decreasing ever so slightly en route to ensure we are following the great circle line between the 2 points.
Paco also made a good point about the 'INS' equipped thing in the question. It's simply the authority's way of telling you you're going to follow a great circle path.
Q 2 ->
For this we need to know how many degrees North we have gone from our initial starting position. 1 degree of longitude is 60nm, so 2950nm is 49.167 degrees. This is 49 degrees and 10 minutes (.167 x 60minutes per degree = 10 minutes), so we have gone north 49 Degrees and 10 minutes.
Our initial starting position was 04 deg 10' S. We are clearly going to end up in the northern hemisphere, so I find it easier to start off at 49 10 N and subtract the southerly starting position to get our overall final position. So 49 10 N - 04 10 S = 45 00 N.
Now I'm going to presume you meant we turned left by 90 degrees because the other answers don't make sense if we were to turn left 99 degrees (we'd be heading back south when all the answer options are either at our current latitude of 45 00 N or higher).
It should be said though that the answer is clear at this point as +TSRA said, as we are clearly going to cross the anti-meridian and transfer from a W to an E longitude our position will be 45 00N XXX XX E, only one answer option suits this.
Anyway, if we're going north and it says we turn left 90 degrees we are going west now. It says 314 km, we need this in NM, dividing any number in km by 1.852 gives you the NM equivalent. So 314 / 1.852 = 170nm. So we move left (changing our longitude) by 170nm. To find our change in longitude in degrees at this particular latitude we're gonna need another formula.
Departure = distance along a rhumb line. Our rhumb line in this instance is the 45 00 N line of latitude.
Dep = Change in longitude (in minutes) x COS Latitude.
We're going to use this formula to find our change in longitude to complete the second part of our coordinates.
We have our departure as 170NM, we have our latitude also as 45N. So ...
170 = Ch Long x COS 45 ... = Ch Long = 170 / .707 = 240 minutes of Longitude. 240 minutes of longitude divided by 60 gives us 4 degrees as being the change of our longitude. So from 17822W we're heading west 4 degrees.
So 178 22 W + 4 degrees = 182 22 W, so we have an excess of 2 degrees and 22 minutes above 180. Simply subtract 2 degrees 22 minutes from 180 degrees to get 177 degrees and 38 minutes ... 177 38 E is the final position.
So 45N 177 38 E is the answer.
Apologies if the very final bit is difficult to imagine, it would be far easier to explain in person and with a globe in hand, it might be difficult to imagine but hopefully you can understand it and it clicks.
This explanation also assumes an understanding of when to add and subtract the conversion angle, having a quick look at flight radar and seeing what paths aircraft take when heading west/north and in south/north hemisphere should help to gain a vision, it's easier to see for long haul flights. This is how I always remembered it and it's drilled it in, others in my class drew out some small diagrams that helped them remember it until it stuck, I'm sure your course materials have something on it.
Q 1 ->
For this we need to use the conversion angle formula.
CA = 1/2 (Change in Longitude x SIN Latitude)
From the question we can establish our longitude will change by 10 degrees and our latitude is 56N
So back to the formula ... CA = 1/2 (10 x SIN 56) ... = 1/2 (8.29) = 4.1 deg ... so 4 degrees.
Now we have to refer back to the question to decide what we're gonna do with this 4 degrees. Once you establish we are heading west along the same line of latitude we can say we are heading 270 degrees. We are also in the northern hemisphere and heading west, this means when departing we are gonna head further north than simply following the rhumb line of the 56N latitude line, this shortens our journey time, we know exactly how much we need to turn north of the 56N line of latitude by using the CA formula as above. So adding 270 + 4 = 274. Again, this will be our heading on departure and will be constantly decreasing ever so slightly en route to ensure we are following the great circle line between the 2 points.
Paco also made a good point about the 'INS' equipped thing in the question. It's simply the authority's way of telling you you're going to follow a great circle path.
Q 2 ->
For this we need to know how many degrees North we have gone from our initial starting position. 1 degree of longitude is 60nm, so 2950nm is 49.167 degrees. This is 49 degrees and 10 minutes (.167 x 60minutes per degree = 10 minutes), so we have gone north 49 Degrees and 10 minutes.
Our initial starting position was 04 deg 10' S. We are clearly going to end up in the northern hemisphere, so I find it easier to start off at 49 10 N and subtract the southerly starting position to get our overall final position. So 49 10 N - 04 10 S = 45 00 N.
Now I'm going to presume you meant we turned left by 90 degrees because the other answers don't make sense if we were to turn left 99 degrees (we'd be heading back south when all the answer options are either at our current latitude of 45 00 N or higher).
It should be said though that the answer is clear at this point as +TSRA said, as we are clearly going to cross the anti-meridian and transfer from a W to an E longitude our position will be 45 00N XXX XX E, only one answer option suits this.
Anyway, if we're going north and it says we turn left 90 degrees we are going west now. It says 314 km, we need this in NM, dividing any number in km by 1.852 gives you the NM equivalent. So 314 / 1.852 = 170nm. So we move left (changing our longitude) by 170nm. To find our change in longitude in degrees at this particular latitude we're gonna need another formula.
Departure = distance along a rhumb line. Our rhumb line in this instance is the 45 00 N line of latitude.
Dep = Change in longitude (in minutes) x COS Latitude.
We're going to use this formula to find our change in longitude to complete the second part of our coordinates.
We have our departure as 170NM, we have our latitude also as 45N. So ...
170 = Ch Long x COS 45 ... = Ch Long = 170 / .707 = 240 minutes of Longitude. 240 minutes of longitude divided by 60 gives us 4 degrees as being the change of our longitude. So from 17822W we're heading west 4 degrees.
So 178 22 W + 4 degrees = 182 22 W, so we have an excess of 2 degrees and 22 minutes above 180. Simply subtract 2 degrees 22 minutes from 180 degrees to get 177 degrees and 38 minutes ... 177 38 E is the final position.
So 45N 177 38 E is the answer.
Apologies if the very final bit is difficult to imagine, it would be far easier to explain in person and with a globe in hand, it might be difficult to imagine but hopefully you can understand it and it clicks.
This explanation also assumes an understanding of when to add and subtract the conversion angle, having a quick look at flight radar and seeing what paths aircraft take when heading west/north and in south/north hemisphere should help to gain a vision, it's easier to see for long haul flights. This is how I always remembered it and it's drilled it in, others in my class drew out some small diagrams that helped them remember it until it stuck, I'm sure your course materials have something on it.
Last edited by Officer Kite; 31st Dec 2017 at 14:25.
Join Date: Oct 2007
Location: Wherever I go, there I am
Age: 43
Posts: 804
Likes: 0
Received 0 Likes
on
0 Posts
The term INS is not necessarily filler - it's often used as code for great circle.
Question everything; including the question itself.
Challenge accepted..
Join Date: Jun 2014
Location: Earth
Posts: 666
Likes: 0
Received 0 Likes
on
0 Posts
Jaair
Are the formulas you have used on the syllabus? That brings back my applied mathematics days from school
Everything on the ATPL syllabus regarding these questions is solvable by the relatively simple conversion angle and departure formulas
Are the formulas you have used on the syllabus? That brings back my applied mathematics days from school
Everything on the ATPL syllabus regarding these questions is solvable by the relatively simple conversion angle and departure formulas
Join Date: Nov 2000
Location: White Waltham, Prestwick & Calgary
Age: 72
Posts: 4,149
Likes: 0
Received 29 Likes
on
14 Posts
We try to avoid the use of formulae as much as possible 
You can answer many questions in Nav (and others) simply by knowing a basic principle and using a diagram.
You can answer many questions in Nav (and others) simply by knowing a basic principle and using a diagram.
Join Date: Jun 2014
Location: Earth
Posts: 666
Likes: 0
Received 0 Likes
on
0 Posts
Ok, but how would you have calculated 4 degrees for example in Q1? Like using the principle of increasing our heading going west in the NH, we also have 278 as an option ... I'm certainly interested 
Join Date: Aug 2013
Location: Spain
Posts: 165
Likes: 0
Received 0 Likes
on
0 Posts
Officer Kite, I have no idea if they're on the syllabus as I'm only PPL and soon to start ATPL (call me crazy for already wanting to attempt ATPL questions). But if I had to guess, I doubt such a formula would be on the syllabus.
Join Date: Jun 2014
Location: Earth
Posts: 666
Likes: 0
Received 0 Likes
on
0 Posts
Upon further inspection there's really not much to it, however it would confuse many I imagine because at first sight it looks a handful. I'm pretty confident it is not on the syllabus otherwise we would have covered it. Although don't put it past easa to come up with a question on it 
Join Date: Nov 2000
Location: White Waltham, Prestwick & Calgary
Age: 72
Posts: 4,149
Likes: 0
Received 29 Likes
on
14 Posts
You don't need to calculate anything really - answers a and b are irrelevant anyway because they are going the wrong way. It's only a short distance so it is most likely to be 274.
Join Date: May 1999
Location: Bristol, England
Age: 65
Posts: 1,804
Received 0 Likes
on
0 Posts
That's true, change of long was only 10 degrees, the maximum the convergency could possibly be was ch long, 10 degrees, and that only at the pole. Max possible conversion angle is therefore 5 degrees, so only one possible answer. Having said that a solution with that sort of detached view would be quite a thing to expect from a student at the very start of their studies, it would be much more likely they would whip through the convergency formula to find a more accurate answer, but with less mental effort.