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Old 1st November 2016 | 09:44
  #901 (permalink)  
 
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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.
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Old 1st November 2016 | 11:29
  #902 (permalink)  
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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.
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Old 1st November 2016 | 14:04
  #903 (permalink)  
 
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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.
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Old 4th December 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?
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Old 5th December 2016 | 15:38
  #905 (permalink)  
 
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From: NY
Don't bother banging your head against those sort of questions.

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.
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Old 6th December 2016 | 10:39
  #906 (permalink)  
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No pressure pattern, but temperature rise, what Jeppesen calls the "new method" of TAS calculation.
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Old 14th December 2016 | 02:32
  #907 (permalink)  
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From: lisboa
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.
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Old 14th December 2016 | 05:50
  #908 (permalink)  
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As with any downgoing wing, the relative airflow comes more from underneath and increases the angle of attack.
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Old 14th December 2016 | 11:54
  #909 (permalink)  
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From: lisboa
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.
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Old 14th December 2016 | 12:56
  #910 (permalink)  
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From: lisboa
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 December 2016 at 16:18.
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Old 14th December 2016 | 13:24
  #911 (permalink)  
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From: lisboa
Paco, I think you are right, but I don´t find any graphic with vectors to see that distribution of forces.
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Old 14th December 2016 | 23:45
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From: Earth
Hi folks,

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

Thanks
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Old 15th December 2016 | 06:21
  #913 (permalink)  
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In the UK, February. As usual, 1500 new questions, and 2000 reviewed.
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Old 19th December 2016 | 19:49
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From: earth
Hi Paco,

Could you give us more information about what will change in the ATPL in 2017.
What do you mean by written questions ?
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Old 21st December 2016 | 23:50
  #915 (permalink)  
 
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From: France
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
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Old 22nd December 2016 | 06:26
  #916 (permalink)  
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From: FL060
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.
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Old 22nd December 2016 | 06:27
  #917 (permalink)  
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An Aviat 67 is an expensive beast, though extremely well built. If you were going to use a slide computer, you won't get a better one - the AFE ARC 2 would be another good choice.

I still use my whizzie after over 40 years - it doesn't need batteries so no embarrassing silences when the calculator dies Also be aware that the electronic ones tend to use the American nautical mile which is a few feet less than the standard one of 6080 feet.

We train with the Jeppesen CR-3 circular flight computer - it is much easier to use than the Dalton/E6B slide thingies, and can be obtained new for around £26 off ebay. As there is a hole through the middle you can put a bit of string through it and use it one handed when you hang it from a suitable place in the cockpit. In our Beavers, there was a convenient hook by the door.

In the absence of the old workbook, which is well out of print, I have done a replacement (PM for details), and we have done an Android app for it as well.
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Old 22nd December 2016 | 09:16
  #918 (permalink)  
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Point of order, Paco. Both the American and British Admiralty nautical miles were 6080ft. The 'international nautical mile' is 1852m, just over 6076ft!
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Old 22nd December 2016 | 10:43
  #919 (permalink)  
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...and EASA, bless them, define a nautical mile as 1852 km, LO 061 01 05 01 in Annex II to ED Decision 2016/008/R.
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Old 22nd December 2016 | 13:14
  #920 (permalink)  
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I'll have to check with the Admiralty manual, but my understanding is that the British version is the one used for calibration and navigation in general, which is correct at 48 degrees of latitude. According to the Dictionary of Military and Associated terms from the US DOD, "The United States has adopted the international nautical mile equal to 1,852 meters or 6,076.11549 feet." That would be at 45 degrees.

That's a loooooong nautical mile, even for EASA!

But who cares - the calculators use 6076 - I'd forgotten the number, but you kindly supplied it!
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