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# ATPL theory questions

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# ATPL theory questions

1st Nov 2016, 09:44

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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

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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

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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

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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

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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

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No pressure pattern, but temperature rise, what Jeppesen calls the "new method" of TAS calculation.
14th Dec 2016, 02:32

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Why a downgoing wing generates more lift without dihedral?

Hi there!

Iīm revising my notes about spins and I donīt find a proper explanation.

Any comment will be very welcome!

Thanks.
14th Dec 2016, 05:50

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As with any downgoing wing, the relative airflow comes more from underneath and increases the angle of attack.
14th Dec 2016, 08:06

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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

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If describing a force instead of saying "higher angle of attack" is horrible, I think weīll have to burn all the books about aerodynamics on earth...

Anyways, if you read the title with attention youīll realize that Iīm 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 donīt see" the vectors involved when it comes to an aircraft without dihedral.
14th Dec 2016, 12:11

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Originally Posted by smthngdffrnt
If describing a force instead of saying "higher angle of attack" is horrible, I think weīll 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...

...youīll realize that Iīm 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

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Thatīs not sarcasm. Itīs 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) wonīt be inclined upwards, increasing its effective angle of attack/lift?

Thatīs 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

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Paco, I think you are right, but I donīt find any graphic with vectors to see that distribution of forces.
14th Dec 2016, 13:41

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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

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I did have, but he's noticed that it's missing and has asked for it back...
14th Dec 2016, 23:45

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Hi folks,

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

Thanks
15th Dec 2016, 06:21

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In the UK, February. As usual, 1500 new questions, and 2000 reviewed.
19th Dec 2016, 19:49

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Hi Paco,

What do you mean by written questions ?
21st Dec 2016, 23:50

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Hello

I would have a question about the computers.
Do they all do the same things ? I have a CRP5 whereas my school recommends an aviat 617.

Are they really used in real life ? (i.e. in flight)
Obviously they are not useful on any modern jetliner, but what about powerful older aircraft ?
I learned tonight most of the computations mine can do, and I was pleasantly surprised with all it can do, and the simplicity of its use (it seemed a bit daunting initially)

Or are they just used during the ATPL exams ?
Is it allowed to have a calculator during the ATPL exams ?
If so, what's the computer for ? (my calculator could easily do everything that the computer does, since it is programmable)

Thank you
22nd Dec 2016, 06:26

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Hello KayPam,

You have to check with your local CAA for which tools you are allowed to bring to the exams. In Germany, you are only allowed to bring to the ATPL exams pens, pencils, ruler, and a flight computer (ie. Aviat 617). They provide notepaper for calculations as well as a REALLY SIMPLE (read: non-programable) caculator. You probably would find the 617 more useful than a CRP5 since it also has MACH numbers on it. Once you pratice with the 617, you have an answer faster than with an electronic calculator esp. wind calculations, PSR & PET are super fast and easy, calculating MACH with IAS and different temperatures is a snap, etc. As for real world application, my IFR FI also flies executive jets and uses his almost daily to double check the caclulations he receives from dispatch.