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ATPL theory questions

Old 13th Jan 2017, 22:40
  #941 (permalink)  
 
Join Date: Dec 2015
Location: France
Posts: 434
Hello
I just received my access to aviationexam databases.

What should we do if the answer to the question is debatable ?
For instance I once had to choose between :
A STAR means standard instrument arrival
B STAR means standard arrival

My book says : STAR means standard (instrument) arrival

I don't care about this particular question, but what should we do in the general case ?
Does this database really contain most of the questions we'll see on the exam day and their correct answer ?

My method will now be to first read the books, and then do the questions (at the end of each book or at the end of each major chapter of that book), dividing my time equally between them : is this a good strategy ?

I would like to get both 90+% and the real knowledge to pass interviews later
And for that I will be working six full months at 25-30 hours per week (have been doing so for already one month)

Maybe subjects should be separated into a "scientific/critical thinking" category (such as flight mechanics) and another "brainless QCM answering" category like air law, and a different method used for each category ?

Thanks for all suggestions, as I hope to be taking my ATPL th exams once in my life, I won't have a trial run.
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Old 14th Jan 2017, 09:17
  #942 (permalink)  
 
Join Date: May 1999
Location: Bristol, England
Age: 60
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I know you said you didn't care but ICAO DOC 8168, Definitions, gives:

"Standard instrument arrival (STAR). A designated instrument flight rule (IFR) arrival route linking a significant point, normally on an ATS route, with a point from which a published instrument approach procedure can be commenced."

So your book is not 100% correct by including the brackets. A small point I agree but even with that information you would answer (a). In the general case you can either ask the database provider or your ATO for clarification or check online. I would not say that this was actually debatable as VFR arrival procedures also exist and would fall under the definition given at (b).
Alex Whittingham is offline  
Old 14th Jan 2017, 15:02
  #943 (permalink)  
 
Join Date: Dec 2015
Location: France
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I do care, however, about how I could learn this without reading the complete billions of pages of ICAO/EASA documentation.

Similar question in this topic :
http://www.pprune.org/professional-p...ml#post9640592
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Old 17th Jan 2017, 18:24
  #944 (permalink)  
 
Join Date: Dec 2005
Location: UK
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Have a look on YouTube. I found a 5 minute video that explained it all. Once I'd watched it a dozen times.
rudestuff is offline  
Old 17th Jan 2017, 18:25
  #945 (permalink)  

 
Join Date: Nov 2000
Location: White Waltham, Prestwick & Calgary
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Simple geometry, isosceles triangles. Treat grid nav like variation.
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Old 18th Jan 2017, 11:37
  #946 (permalink)  
 
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Your floor will be littered with sheets of A4 paper covered in circles! Try to get 6 each side - think of the trees.
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Old 18th Jan 2017, 16:59
  #947 (permalink)  
 
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It's easy if you keep in mind there are only 180 degrees in a triangle, so if you know the value of two angles you can easily find the third one. And it's even easier if you consider that you count angles always in a clockwise direction relative to the North being considered—be it Grid North, True North, Compass North or Magnetic North.

Simple, not precise sketches will do it for you, keep it simple. And on some occasions if you draw the sketch accurately with a divider and protractor you'll get to the solution without doing any fancy calculation on your calculator. While I was quite worried about Gen Nav at the beginning of the syllabus, it turned out to be one of my favourite subjects.
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Old 18th Jan 2017, 17:33
  #948 (permalink)  
 
Join Date: Mar 2013
Location: EU
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360 degree protractor. I would say 8 times out of 10 from what I remember if you drew a circle round the protractor and put the points on accurately you could get the answer from the multi choice options within a few seconds.
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Old 21st Jan 2017, 18:39
  #949 (permalink)  
 
Join Date: Dec 2015
Location: France
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Hello,
I failed to see that in the following question, the gyro north was at 0 initially ! I thought the gyro north was at the real north initially (how stupid am I?)
https://i.gyazo.com/6c17d40997881b47...31be630be9.png

So I started to make a huge fuss about this, including a polar navigation sketch in Google Earth.
Let's not loose it, as it took a bit of time to make :
https://i.gyazo.com/ce5e898881efc970...187450823b.jpg

If anyone would like my explanation as to how the picture is made and what the reasoning behind it is, do not hesitate to ask further clarification.
KayPam is offline  
Old 23rd Jan 2017, 09:07
  #950 (permalink)  
 
Join Date: May 2016
Location: UK
Posts: 4
XXXX 251020Z 24005KT 5000 NSC 05/03 Q1022 R99/3101//

Can anyone decode, "R99/3101//"?

Thanks
iFunFlyer is offline  
Old 25th Jan 2017, 20:17
  #951 (permalink)  
 
Join Date: Aug 2012
Location: Greece
Age: 38
Posts: 22
Question on operational procedures

Hello!
I was wondering if someone can help me on this question:

You plan to fly from point A (60oN 010oE) to point B (60oN 020oE). The gyro North of the gyro compass, assumed to be operating perfectly, with no rate correction device, is aligned with the true North of point A. The constant gyro heading to be followed when starting from A given that the flight time scheduled is 1h30 min with a zero wind, is equal to:

a)66
b)80
c)76
d)85

Questions:
1)Do we have to draw one of these known diagrams with a circle and grid north or just a just two parallels (010oE and 020oE) are enough?
2)How the solution is affected by the fact that at point A the gyro north is aligned with true north?
3)The two points (A and B) have the same latitude.That means that in order to fly from A to B we have to fly constant TRUE track of 090o?This is a rhumb line?

I would be grateful to anybody who has the will and time to spend some time on this question.

Thanks for reading!
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Old 25th Jan 2017, 21:41
  #952 (permalink)  
 
Join Date: May 1999
Location: Bristol, England
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1) no
2) It's a straight line track, ie a great circle, the question is effectively asking you the initial great circle track A to B. Given that the initial true track = the initial gyro track, if you hold the initial gyro track on the gyro compass it will get you from A to B, true track will change as the true north reference changes (that's convergency in action) but the gyro will not. This is the strength of gyro navigation in high latitudes.
3) 090 deg true would be a rhumb line but this is not a straight line track, and therefore only relevant in calculating the great circle.

Calculate convergency, conv = ch long sin mean lat = 10 sin 60 (ish) = 8.6 deg,
Conversion angle = 1/2 convergency = 4.3 deg
Rhumb line = 090 deg, therefore initial great circle track = 090 - 4.3 = 085.7 deg
initial great circle track 085.7 deg true therefore grid track = 085.7 deg.
QED
....a bit heavy, though, this is really General Nav, and quite foxy for that.
Alex Whittingham is offline  
Old 25th Jan 2017, 22:16
  #953 (permalink)  
 
Join Date: Dec 2015
Location: France
Posts: 434
Hi
You may (or may not) be interested in plotting circles such as the ones shown on the second link in my post below :
http://www.pprune.org/professional-p...ml#post9648882 (the figure was not modified to fit your case, just consider the 62 parallel and its intersection with the 30W and the 10W meridians)

Then you will be able to see for yourself exactly how this all works in Google Earth

You will be able to see that halfway between, your gyro track of 85.7 gives you a true track of 90 and that your true track upon arrival will be 94.3
Your average true track will be exactly 90

Hope this clarified the situation as opposed to making it more confused.
KayPam is offline  
Old 25th Jan 2017, 22:16
  #954 (permalink)  
 
Join Date: Aug 2012
Location: Greece
Age: 38
Posts: 22
Thanks for the answer!

The truth is that I am confused because I have seen other solutions that comprise astronomic and transport precession,they get average of them and substract from the rhumb line.
It gives another result,totally different.

Why so different?
If the instructors can't agree on a solution that gives the same result,why do they expect it from the students?
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Old 25th Jan 2017, 22:46
  #955 (permalink)  
 
Join Date: Dec 2015
Location: France
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And now you got me confused !

It indeed seems like time required to travel should be taken into account.
Except of course if something in the question precludes it. Which does not seem to be the case here.

New answer would be 76

Alex, please ?
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Old 26th Jan 2017, 06:09
  #956 (permalink)  

 
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There's no rate correction device, and the flight time is 90 minutes, or 1.5 hours, meaning 200 kts.....

076, based on -13 (Earth rate in the NH*) and -1 (Transport Wander heading East)**. 90 - 14 = 76.

*15 x .866
**200 x tan lat .32/60

Last edited by paco; 26th Jan 2017 at 06:41.
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Old 26th Jan 2017, 07:42
  #957 (permalink)  
 
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You're right that it is more complex than I thought, I took no account of the earth's rotation. Please disregard my answer above, I will think about Paco's solution.
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Old 26th Jan 2017, 09:02
  #958 (permalink)  

 
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You're right, though - what the h*ck is it doing in Ops?
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Old 26th Jan 2017, 09:22
  #959 (permalink)  
 
Join Date: May 1999
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My early thoughts are that, if the initial GC track is 085.7 deg true, a starting track of 076 deg G (which is necessarily also 076 deg true) will not fly you either a great circle or rhumb line.
Alex Whittingham is offline  
Old 26th Jan 2017, 13:40
  #960 (permalink)  
 
Join Date: May 1999
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OK here we go, for gyro experts to comment on. Paco, I get TW of 9 degrees not 1 degree, I don't understand the 0.32 in your formula. I think if you are going to express TW as a rate the formula is (east west groundspeed*tan lat/60)*hours flown which gives the same answer as working it out as convergency.

Solution:

Calculate convergency, conv = ch long sin mean lat = 10 sin 60 = 8.6 deg,
Conversion angle = 1/2 convergency = 4.3 deg
Rhumb line = 090 deg T, therefore initial great circle track = 090 - 4.3 = 085.7 deg T
This will also be initial gyro heading as they are aligned
Final GC track will be 090 + 4.3 = 094.3 deg T

Assuming first of all for the sake of argument that the great circle track is followed by reference to other instruments (INS?) and the gyro is compensated for latitude the total gyro drift will be:
ER + LN + TW
= ([-15 sin 60]*1.5) + ([+15 sin 60]*1.5) + (-10 sin 60)
= (-13) + (+13) + (-9)
= -9
So the final gyro heading will be 094.3 – 9 = 085.3

Now modifying the above so that we continue to fly the GC track by other means but the gyro is now not compensated for latitude, the total gyro drift will be
ER + LN + TW
= ([-15 sin 60]*1.5) + (0) + (-10 sin 60)
= (-13) + (0) + (-9)
= -22
So the final gyro heading in this case will be 094.3 – 22 = 072.3

By which calculations we can see that flying a constant gyro heading/track is not going to follow the shortest (great circle) route. Is it true that flying the ‘average gyro heading/track’ will still get us from A to B by a curved track lying to the north of the great circle? I think so.

If so the ‘average’ to approximate to the GC track without latitude nut correction is (085.7+072.3)/2 = 079 deg

I can't immediately see logically why any calculations should be based on gyro corrections applied to the rhumb line track as we are after the shortest path but, interestingly, if you follow the same argument through and try and fly a rhumb line track by other means the initial true and gyro headings would be 090, the final true heading would be 090, gyro errors would be the same -22 so final gyro heading would be 068, and the average gyro track also 079. This is because the average great circle track is 090, and by averaging out initial and final gyro headings on a great circle track we average out the 'great circle elements' and arrive at the same answer. At one level not surprising because we would not expect to find two constant gyro headings/tracks between two points.

Last edited by Alex Whittingham; 26th Jan 2017 at 16:02.
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