# ATPL theory questions

Join Date: Jun 2012

Location: -

Posts: 1,175

Likes: 0

Received 0 Likes
on
0 Posts

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.

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.

Join Date: Mar 2013

Location: EU

Posts: 497

Likes: 0

Received 0 Likes
on
0 Posts

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.

Join Date: Dec 2015

Location: France

Posts: 507

Likes: 0

Received 0 Likes
on
0 Posts

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.

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.

Join Date: Aug 2012

Location: Jupiter

Posts: 28

Likes: 0

Received 0 Likes
on
0 Posts

**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!

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!

Join Date: May 1999

Location: Bristol, England

Age: 65

Posts: 1,808

Received 0 Likes
on
0 Posts

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.

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.

Join Date: Dec 2015

Location: France

Posts: 507

Likes: 0

Received 0 Likes
on
0 Posts

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.

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.

Join Date: Aug 2012

Location: Jupiter

Posts: 28

Likes: 0

Received 0 Likes
on
0 Posts

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?

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?

Join Date: Dec 2015

Location: France

Posts: 507

Likes: 0

Received 0 Likes
on
0 Posts

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 ?

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 ?

Join Date: Nov 2000

Location: White Waltham, Prestwick & Calgary

Age: 72

Posts: 4,164

Likes: 0

Received 29 Likes
on
14 Posts

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

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.*

Join Date: May 1999

Location: Bristol, England

Age: 65

Posts: 1,808

Received 0 Likes
on
0 Posts

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.

Join Date: May 1999

Location: Bristol, England

Age: 65

Posts: 1,808

Received 0 Likes
on
0 Posts

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.

Join Date: May 1999

Location: Bristol, England

Age: 65

Posts: 1,808

Received 0 Likes
on
0 Posts

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

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

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

I can't immediately see

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 beER + 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 degI 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.*

Join Date: Nov 2000

Location: White Waltham, Prestwick & Calgary

Age: 72

Posts: 4,164

Likes: 0

Received 29 Likes
on
14 Posts

the .32 was supposed to be the tan of 60, at least that's what the internet site I used said .

Multiply by the groundspeed and divide by 60 was the formula for TW.

60

However, having used a real calculator this time , the tan lat is 1.73, so that ends up with 5.77 or 6 degrees for government work

As it is minus because you are going E, apply to the -13 for the apparent wander to get -19, subtract from 90 to get 81, answer B

Multiply by the groundspeed and divide by 60 was the formula for TW.

__G/S x tan lat__60

However, having used a real calculator this time , the tan lat is 1.73, so that ends up with 5.77 or 6 degrees for government work

As it is minus because you are going E, apply to the -13 for the apparent wander to get -19, subtract from 90 to get 81, answer B

Join Date: May 1999

Location: Bristol, England

Age: 65

Posts: 1,808

Received 0 Likes
on
0 Posts

its a correction per hour, so you need to multiply 5.77 x 1.5 = 8.66. Same as the convergency, because its the convergency that causes TW.

I'm guessing you had put your calculator down by the time you said 90 - 19 = 81!

I'm guessing you had put your calculator down by the time you said 90 - 19 = 81!