I used to think it was 30 seconds, but according to ICAO Annex 10, Volume 2, Chapter 5, 10 seconds is correct:Interestingly, this only applies to calling aeronautical stations, not aircraft stations.

I was guessing 30 seconds, but then i counted 10 seconds and realized this seems as an eternity when it comes to quick decisions. Thanks for the info. :D

Now since we're at this topic, you know what's the difference when it comes to a ground aeronautical station? (since you mentioned it :p)

Sorry, I just realise I confused the terms 'aeronautical station', which - in ICAO terminology - means ground station, and 'aircraft station'. The respective definitions, as given in ICAO Annex 10, Volume 2, Chapter 1, are:

Airport Facility Directory Question
 

Hello. Sorry for the disturbance, people...

This question on the FAA ATPL written examination study guide;

I look into the A/FD excerpt and see the following under COMMUNICATIONS;


I would have said, based on the arrival, that you are inbound on radial 068 of LAX VOR, therefore the answer should be 128.5 MHz (between LAX radials 045 - 089). The guide says the answer is 124.5 MHz.

Some background now, please, before I get pitched into about not understanding the A/FD. I am not familiar with the A/FD format, really, and am preparing for the exam based on a company requirement (I work for a South American based airline). On what premise is the provided answer the correct one???

Thanks!

I'm not familiar with this format. I haven't come across anything similar in my relatively limited European experience.

To hazard a guess, if the A/FD excerpt is referring to your inbound track as opposed to what radial you're on (like the MSA is stylized on jepp plates) then if you are inbound on radial 068, your course or nil wind track would be 248 which lies between 225 and 044, thus giving 124.5.

Okay, thanks. I imagine that must be it. It is weird, though. I have looked for "guides" to interpreting the A/FD online, but cannot find anything.

Argh! Sorry guys,

I did not initially notice that there was a sticky thread for exactly my sort of query, for which I posted a separate thread in this section of the forum. For the record, here's a link to it;

A/FD Question

And if anyone can further offer any explanation, I would be most appreciative, thank you.

the drag coefficient at constant angle of attack

A:starts to increase rapidly above Mcrit
B:starts to decrease rapidly above the drag divergence mach number
C:increases only when Mcrit is above unity
D:starts to increase rapidly above the drag divergence Mach number

If you do a google search for "Drag Divergence Mach Number" you will find lots of material including diagrams on this subject.

The WIKIPEDIA entry starts with the following text:

The drag divergence Mach number (not to be confused with critical Mach number) is the Mach number at which the aerodynamic drag on an airfoil or airframe begins to increase rapidly as the Mach number continues to increase.[1] This increase can cause the drag coefficient to rise to more than ten times its low speed value.

Hi all,

I'd need some help with the following question:

Departure A (25°N 175°W) on 7 January at 1423 LMT. Difference UTC and ST(a) is 11 hr.
Destination B (15°N 155°E). Difference UTC and ST(b) is 10 hr.
Distance along the great circle between A and B is 1790 NM. Average head wind is 19 kt, average TAS 400 kt. Calculate time (standard) and date of arrival at B.

So, I came up with 16:05 (8 Jan) but according to the QDB the correct answer is 16:45 (8 Jan)

Here are my steps how I came up with my answer:

1. Trip time = Distance / Ground Speed = 1790 NM / 381 = 4 hr 42 min
2. Convert from LMT to UTC (departure airport) 1423 + 11 = 01:23 (8 Jan)
3. from UTC to LMT (destination) 01:23 + 10 = 11:23
4. add trip time: 11:23 + 4:42 = 16:05 (8 Jan)

Many thanks in advance!

Hi transsonic,

Unfortunately you are confusing LMT and Standard time. The departure time is given in LMT, to convert to UT use the arc to time tables. Thus your step 2 onwards should be:

2. Convert from LMT to UTC (departure airport) 1423 + 11:40 = 02:03 (8 Jan)
3. from UTC to ST (destination) 02:03 + 10 = 12:03
4. add trip time: 12:03 + 4:42 = 16:45 (8 Jan)

Hi Alex,

thank you very much! I see, I should have divided 175°W by 15°/hr which gives me 11 hr 40 min! I was a bite confused because the answer I came up with initially (16:05) wasn't even listed among the possible answers given by this question.

Hi there,

it's me again. I'd need some help with the following question, no clue how to get this figured out.

AT 0020 UTC aircraft is crossing the 310° radial at 40 NM of a VOR/DME.
AT 0035 UTC the radial is 040° and DME distance is 40NM.
Magnetic Variation is zero. What is the true track and ground speed?

correct answer would be 085° - 226 kts


Many thanks in advance!

Heading and Speed Calculations
 

Using my very basic VOR Distance/Time formulas, I managed to get as close as possible to the "Correct Answer" you provided.

First we can conclude that the Aircraft has flown from NW to NE and as it has flown across 90 degrees of Radials (310-040), we can say the aircraft was almost perfectly flying East (090). Its a perfect Isosceles triangle.

First Example: We understand from IFR ground school that a scale of 1 dot is = 200 feet when 1 nm from the station. So 1 degree at 1 nm = 100 feet.
At 40nm 1 degree would be 4000 feet (40nmX100feet).
The airplane flew through 90 Degrees of Radials so 90X4000 = 360,000 feet.
1nm = 6076 feet. Therefore, the Distance flown from the 310 radial to the 040 is 59.24nm(360,000/6076).
Now we can say that in 15 mins (0020-0035) the airplane covered a distance of 59.24nm. The aircraft must be traveling at 236 knots.

Second Example:

60 x minutes flown between
bearing change
Time to station = ----------------------------------
degrees of bearing change

X = 60 x 15
--------
90
X is 10 minutes: Therefore the aircraft will cover 40nm to the station in 10 minutes. This gives a speed of 240 knots

TAS x minutes flown
Distance to station = ------------------------------
degrees of bearing change

Now 240knots x 15
-----------------
90
= 40nm from station.

I guess I can rest my case to say :D:D:D problems solved. At least to the nearest "Correct Answer".

It's all about the properties of isosceles triangles, angles & Pythagoras. However the easiest way is to draw a SCALE diagram - answer then will leap out at you.

Many thanks for your help, especially RVR400, it's very much appreciated!

This is one of these absurd questions the QDB contains. Personally I doubt that one (I) will recall all the steps involved to come up with the right (close) answer on exam day, it's probably just recognizing the question and hopefully the correct answer, if one is lucky.

Once again, many thanks for your explanations :ok:

Trans - first sketch the problem. You have been given some good stuff to start. Two radials, two DME fixes and 15 minutes (or 1/4 of an hour). First the track. Half way, you will be crossing the 355 radial at 90 degrees, so add the two to get 085. Now the distance. Again, at the halfway point, you will be on the 355 radial, having flown from the 310 radial. You have covered 45 degrees. So Cos 40 (0.7071) x 40 = 28.28 x 2 (you are halfway) = 56.57 nm x 4 ( is was a quarter of an hour) = 226 Kts.

Many thanks PM, it's very much appreciated!

GNAV - Compressibility Factor
 

I'd need some help with the following question:

Aircraft is flying at FL 350 with Mach 0.878 OAT = ISA + 4°C. Compressibility Factor is 0.939. Calculate TAS.

Correct answer would be 510 kts.


Her are my steps, but my result doesn't even come close to the answer given.

1. determine OAT at FL 350 = - 51°C
2. calculate LSS = 0.878 (Mach) x 38.9 (square route out of) 222 = 508 kts
3. 508 kts x 0.939 (C. Factor) = 478 TAS

510 kts is correct.

You do not have to account for compressibility errors for Mach Nos.

The Machmeter only has instrument & pressure errors not compressibility, whereas the ASI has instrument, pressure, compressibility & density errors.

Hi Richard,

thank you very much for your explanation, it's very much appreciated! Good to know that step 3. is not necessary, really gave me some headache.

Hello,

this question is found in the Principles of Flight questions in the exams:

Two identical aeroplanes A and B, with the same mass, are flying steady level co-ordinated 20 degree bank turns. If the TAS of A is 130 kt and the TAS of B is 200 kt:

Possible answers:

the turn radius of A is greater than that of B.
the load factor of A is greater than that of B.
the rate of turn of A is greater that of B. (correct)
the lift coefficient of A is less thant that of B.

Is it, just theoretically, possible, that the last question could be also correct, since the question doesn´t state anything about same aircraft configuration (flaps/spoilers) or same air density?

Pressure height Q from Aviationtire.com
 

Find the airfield pressure height given:
Elevation 397ft
QNH 1023 hpa
OAT +22 C

answers
A) 97ft
B) 397 ft
C) 127 ft
D) 291 Ft


...and they saying that right answer will be C) 127 ft.

Just how? If someone could put me on the track with this. My guess is 97ft.

:*

They used 27' per millibar instead of 30 I guess?

Identical aeroplanes with same mass and same AOB suggests that they are both producing the same amount of lift.

Lift = CL 1/2Rho TAS squared

Aircraft A is flying at a lower TAS, so to generate the same lift as aircraft B with the same air density it would require a greater CL. This is not an option in the question.

It would be possible for aircraft A to generate the same lift with a lower CL than aircraft B, but to achieve this it would need a value of Rho that was greater than that for aircraft B. Nothing in the questions suggests that this is the case, so option D is unlikely to be the correct answer. You would be unwise to select this option in the exam.

But there is certainly enough information to indicate that option C is the correct answer.

Da-a,

and in those expensive JAA ATPL books they teaching use 30ft per millibar. So why there is a choice 97ft which you will get when use 30ft / millibar.

Go figure!

Yea. I've seen in many questions that they actually state "use 30 feet per millibar" or "use 27 feet per millibar". To have two answers that are correct for each of these is a bit evil imo. I guess 27 is the more correct answer, but it depends on where you are in the atmosphere.

96 x (absolute temperature (273K + actual temperature) / altitude pressure (millibars))

So at sea level in ISA conditions that's 27.3 feet.

So if you have the choice, and it's correct to use 27 feet, I guess that would be the "more correct" answer (at low levels). At 18,000 feet it's closer to 50 feet per millibar


Someone correct me if I'm wrong

Unless otherwise specified, we teach 27 feet for Met, 30 feet for everything else

ATPL theory questions
 

Don't forget you can now comment on ATPL questions. I recently had one of mine upgraded presumably after they read one oft comments on it. So if you put in that your answer is based on 30 or 27 then you may find they credit it regardless.

And the charge is £69 for the privilege.....

Wind Component for Flight Planning
 

If a "wind component is +40kt" would this refer to a headwind or tailwind, the question doesnt specify.

I've assumed tailwind as your GS would increase in this case.

In Navigation headwinds have a negative effect because they decrease the ground speed and this decreases the ground distance.

In Aircraft Performance (particularly take-off and landing performance) headwinds have a positive effect because they decrease the ground speed, which decreases the ground distances required.

It would help if we knew the whole question.

Thanks Keith. It's a nav question (flight planning)

Ref CAP697 MEP 3.6

Cruise PA 18000ft OAT +5
Airfield PA 4000ft OAT +20
Wind component in descent +40kt

What are the fuel, time and distance (NGM)

I got fuel 8-2 = 6USG, time 19-4 = 15mins, distance 52-11 + 15/60*40 = 51NGM

i.e. cover more distance in same time due to tailwind.

I agree with your figures.

Worth noting that the exams are no longer giving you reference numbers e.g. 3.6. You are expected with the information in the question and a bit of ingenuity get the correct graph/table. Watch out on MRJT.

I believe the qb has been updated today. Can anyone else back up the statement?

Since they update very regularly, it is extremely important to study also the books very well. That saved me while I was doing my ATPL Theory exams.

Just clicking through the questions will not help you. It is just a tool to practice your knowledge.

When I did my exams, they were also updating many subjects. Seen many questions I ve never seen before in QB. But my book knowledge made me passing first attempts. Never failed any of them.

Good for you, good for you! :D

I read the books AND study using the QB, as it is there as an available tool for us.

I just discovered that it was quite a big change in the total of questions in it (atp), so i figured someone else using bristol ground school would see it as well.

Gen Nav Mercator Projections
 

I'm working through a ProPilot workbook so I can't check my answer to this (as they aren't provided) but just wanted to check I'm going about it the right way

The total length of the 53N parallel of latitude on a direct Mercator chart is 133cm. What is the approximate scale of the chart at latitude 30S.

My working:

Circumference of earth at 53N = Pi(12732*cos53) = 7662km
Scale at 53N = 1 over 766,200,000/133 = 1/5,760,902
Scale at 30S = cos30/cos53 * 5,760,902 = 1/8,229,860

Can anyone confirm if I've done this correctly?

Thanks.

Might be best to approach your FTO directly, you are on the right lines but this should help. This JAA exam question was asked years ago and does cause some problems - you are not alone.

Scale is Earth Distance(ED) / Chart Length(CL) (as a ratio) saves all this 1 over rubbish. You must work in the same units Km/cm NOT Nm & cm.

Next thing is to spot it's a Mercator question so the 133cm CL applies at ALL latitudes so you can dismiss the 53N bit and go straight for ED at 30s. This brings in "departure". Dept (NM) = Ch long (mins) x cos lat.

So 360 * 60 * cos 30 = 18706 nm, however CL is in cm so convert 18706 into km by * 1.852 = 34643.512 km (worth remembering 100,000 cm in km) so * by 100,000 = 3464351200 then divide by the CL of 133 gives answer of 26,047,753.

The original exam answer was 1:25,000,000 which is close enough.

NB. Several questions on scale where you have to find ED by departure before you can do anything else.

speed instability
 

Good evening to you all !

Could any of you enlight me as to why a swept wing aircraft would be more speed instable than a classic straight winged ?

Thanks for your help :)