ATPL theory questions

[hvogt]

I used to think it was 30 seconds, but according to ICAO Annex 10, Volume 2, Chapter 5, 10 seconds is correct:

5.1.5 Recommendation.—After a call has been made to the aeronautical station, a period of at least 10 seconds should elapse before a second call is made. This should eliminate unnecessary transmissions while the aeronautical station is getting ready to reply to the initial call.

Interestingly, this only applies to calling aeronautical stations, not aircraft stations.

[teddz]

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.

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 )

[hvogt]

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:

Aeronautical station [...] A land station in the aeronautical mobile service. In certain instances, an aeronautical station may be located, for example, on board ship or on a platform at sea.

Aircraft station [...] A mobile station in the aeronautical mobile service, other than a survival craft station, located on board an aircraft.

[Gallinazo]

Airport Facility Directory Question

Hello. Sorry for the disturbance, people...

This question on the FAA ATPL written examination study guide;

Which approach control frequency is indicated for the TNP.DOWNE3 Arrival with LAX as the destination?

128.5 MHz
124.9 MHz
124.5 MHz

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

APP CON: 128.5 (045º - 089º), 124.9 (090º - 224º), 124.5 (225º - 044º)

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!

[OhNoCB]

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.

[Gallinazo]

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.

[Gallinazo]

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.

[Ahmedsalo]

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

[keith williams]

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.

[Transsonic2000]

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!

[Alex Whittingham]

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)

[Transsonic2000]

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.

[Transsonic2000]

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!

[RVR400]

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 problems solved. At least to the nearest "Correct Answer".

[RichardH]

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.

[Transsonic2000]

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

[Piltdown Man]

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.

[Transsonic2000]

Many thanks PM, it's very much appreciated!

[Transsonic2000]

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

[RichardH]

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.