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# ATPL theory questions

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# ATPL theory questions

1st Nov 2016, 09:36

Join Date: May 1999
Location: Bristol, England
Age: 60
Posts: 1,472
Just to double clarify the answer to your last sentence, RedBull, there is no separate scale printed on the chart because you are expected to know that one minute of arc on a great circle is 1NM. Meridians are great circles so if you measure one degree of latitude change along a meridian you are measuring 60NM and, as Paco says, 300NM is 5 degrees of latitude change.
1st Nov 2016, 09:44

Join Date: Jun 2012
Location: -
Posts: 1,176
I printed the annex paper of course.
I did some research and they key to solving this question was to assume 1cm=100NM.
1st Nov 2016, 11:29

Join Date: May 1999
Location: Bristol, England
Age: 60
Posts: 1,472
If that worked it was only by chance. There was no scale given other than the meridian. The explanation was:

(Refer to Diagram)

From the North Pole on the diagram the aircraft flies down the 110°East meridian.

The meridians are shown every 10° of longitude.

The distance travelled is 480 NM which is equal to 8° of latitude.

Use the latitude scale to measure the distance flown from the pole down the 110° East meridian.

The aircraft is now at 82°N 110°E.

At this point draw in grid North parallel to the Greenwich meridian and plot a track of 154° Grid from 82°N 110°E.

The aircraft travels 300 NM along this track which is equal to 5° of latitude.

Use the latitude scale to measure this 300 NM, plot it onto the chart and you should find the aircraft will be at or close to 80°N 080°E.
1st Nov 2016, 14:04

Join Date: Jun 2012
Location: -
Posts: 1,176
I couldn't understand just the last row of the explanation as that was the final point where I was stuck, however now it's crystal clear.
4th Dec 2016, 22:41

Join Date: Jun 2012
Location: -
Posts: 1,176
Just a couple of quick questions.

I am switching to a Jeppesen CR flight computer and noticed in the manual that it is possible also to calculate so called pressure patterns problems. Since I have not find a lot about it in my ATPL notes, am I right in assuming it is not part of the syllabus?

Second question is, are temp rise questions part of the syllabus?
5th Dec 2016, 15:38

Join Date: Nov 2016
Location: NY
Posts: 5

Some of them are just written poorly and are quite vague in what they ask for.
There are some ultimately you'll just have to remember.
6th Dec 2016, 10:39

Join Date: Nov 2000
Location: White Waltham, Prestwick & Calgary
Age: 67
Posts: 3,749
No pressure pattern, but temperature rise, what Jeppesen calls the "new method" of TAS calculation.
14th Dec 2016, 02:32

Join Date: Jul 2016
Location: lisboa
Posts: 62
Why a downgoing wing generates more lift without dihedral?

Hi there!

I´m revising my notes about spins and I don´t find a proper explanation.

Any comment will be very welcome!

Thanks.
14th Dec 2016, 05:50

Join Date: Nov 2000
Location: White Waltham, Prestwick & Calgary
Age: 67
Posts: 3,749
As with any downgoing wing, the relative airflow comes more from underneath and increases the angle of attack.
14th Dec 2016, 08:06

Join Date: Nov 2015
Location: Farnham, Surrey
Posts: 1,191
You're no being clear, but if the question is what I think it is:

Dihederal increases the angle of attack (AoA)on the outside wing in yawed flight. So if an aeroplane with dihederal is yawed to the left (left pedal) the AoA of the right wing increases, and the AoA of the left wing decreases - producing a roll moment in the same direction as the yaw. So in principle if an aeroplane is yawed and rolled in the same direction there will be "less lift" (horrible way of describing it) on the down-going wing than there would be if the dihederal wasn't present.

Now you mentioned spinning, but that's a different skillet of sea-bass because in a spinning condition the "down-going wing" is mostly stalled - the loss of lift and increased drag being what produces the autorotating couple. But I'm sure this isn't what you're asking, so could you clarify the question?
14th Dec 2016, 11:54

Join Date: Jul 2016
Location: lisboa
Posts: 62
If describing a force instead of saying "higher angle of attack" is horrible, I think we´ll have to burn all the books about aerodynamics on earth...

Anyways, if you read the title with attention you´ll realize that I´m talking about wings without geometric dihedral. In other words, in case of sideslip, why does the downgoing wing have a higher angle of attack?

I understand this scenario when we have an aircraft with physical dihedral, however, It seems that "I don´t see" the vectors involved when it comes to an aircraft without dihedral.
14th Dec 2016, 12:11

Join Date: Nov 2015
Location: Farnham, Surrey
Posts: 1,191
Originally Posted by smthngdffrnt
If describing a force instead of saying "higher angle of attack" is horrible, I think we´ll have to burn all the books about aerodynamics on earth...
It's a horrible way of describing it because there isn't "less lift" - there is just a spanwise redistrubution of the same amount of lift.

Anyways, if you read the title with attention...
Yes, being sarcastic with people who bother to try to answer vaguely-expressed questions always motivates further responses, doesn't it...

...you´ll realize that I´m talking about wings without geometric dihedral. In other words, in case of sideslip, why does the downgoing wing have a higher angle of attack?
It doesn't. Any yaw-roll coupling in aeroplanes with no geometric dihederal comes from other sources - wing-sweep, differential blanketting of inboard wing sections and secondary roll moment of the fin being the most common ones AIUI.
14th Dec 2016, 12:56

Join Date: Jul 2016
Location: lisboa
Posts: 62
That´s not sarcasm. It´s just an objective fact as you were talking about dihedral and the title is clear (without dihedral). On the other hand, you say vaguely-expressed, I say briefly-expressed -Paco got it at first try-.

As far as the technical content of your answer, do you affirm that the relative airflow (downgoing wing in a sideslip) won´t be inclined upwards, increasing its effective angle of attack/lift?

That´s what the books say literally and you say that the amount of lift is the same. Now I understand nothing.

Thank you, PDR1.

Last edited by smthngdffrnt; 14th Dec 2016 at 16:18.
14th Dec 2016, 13:24

Join Date: Jul 2016
Location: lisboa
Posts: 62
Paco, I think you are right, but I don´t find any graphic with vectors to see that distribution of forces.
14th Dec 2016, 13:41

Join Date: Nov 2015
Location: Farnham, Surrey
Posts: 1,191
Not sure why I'm bothering, but anyway.

If flying in balanced straight and level flight with zero yaw and an angle of attack alpha is subjected to a pure yaw (no roll or pitch secondary effects of the yaw control) through an angle beta then the angle of attack of the wing (the whole wing, not one side or the other) will reduce by a factor proportional to cos (beta).

In a real aeroplane the resulting reduction of lift coefficient will make lift less than weight, so the aeroplane will accelerate downwards until it achieves a steady state condition where the angle of descent is equal to alpha * [1-cos(beta)].

At least I think that's it - trying to visualise it in my head because I'm posting this with my phone and don't have a whiteboard handy to sketch the diagram.

But I cannot see any situation where it would be different for the two sides of a wing with no dihederal.
14th Dec 2016, 14:23

Join Date: Nov 2015
Location: Farnham, Surrey
Posts: 1,191
I did have, but he's noticed that it's missing and has asked for it back...
14th Dec 2016, 16:15

Join Date: Jul 2016
Location: lisboa
Posts: 62
Council Van, I´m just "contending" at the same level. I mean, I´m not the one who describes something stated in books as horrible -in public-. Anyways, if you are taking this brainstorm personally, feel free to delete it. "I will find myself" at the airport.

Last edited by smthngdffrnt; 14th Dec 2016 at 17:30.
14th Dec 2016, 20:28

Join Date: Jul 2016
Location: lisboa
Posts: 62
I do agree with you. I don´t think saying "that quotation is horrible" is a very good way to approach and communicate with people, let alone if it is used -literally- in some well-known books of aerodynamics.

I don´t really know if it´s a language issue, what I know is that some people don´t know how to apply their own invaluable advices -in both directions-.

Take care and thank you for your gratitude.

My most sincere gratitude for the technical part, PDR1.

Last edited by smthngdffrnt; 14th Dec 2016 at 20:56.
14th Dec 2016, 23:45

Join Date: Jun 2012
Location: Earth
Posts: 71
Hi folks,

someone heard about the change of the exams in April ? Apparently 20% will be written?

Thanks
15th Dec 2016, 06:21

Join Date: Nov 2000
Location: White Waltham, Prestwick & Calgary
Age: 67
Posts: 3,749
In the UK, February. As usual, 1500 new questions, and 2000 reviewed.