# ATPL theory questions

Join Date: Dec 2006

Location: Hamburg

Age: 46

Posts: 433

Likes: 0

Received 0 Likes
on
0 Posts

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

Interestingly, this only applies to calling aeronautical stations, not aircraft stations.
**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.**Last edited by hvogt; 4th Jul 2014 at 11:47. Reason: see my post below*

Join Date: Jul 2014

Location: Romania

Posts: 2

Likes: 0

Received 0 Likes
on
0 Posts

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 )

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 )

Join Date: Dec 2006

Location: Hamburg

Age: 46

Posts: 433

Likes: 0

Received 0 Likes
on
0 Posts

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.

Join Date: Aug 2012

Location: Quito, Ecuador

Age: 56

Posts: 18

Likes: 0

Received 0 Likes
on
0 Posts

**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!

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*

**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!

Join Date: Mar 2013

Location: EU

Posts: 497

Likes: 0

Received 0 Likes
on
0 Posts

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.

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.

Join Date: Aug 2012

Location: Quito, Ecuador

Age: 56

Posts: 18

Likes: 0

Received 0 Likes
on
0 Posts

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.

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.

Join Date: Jun 2013

Location: Cairo

Posts: 4

Likes: 0

Received 0 Likes
on
0 Posts

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

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

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.

Join Date: Jan 2006

Location: Europe

Posts: 404

Likes: 0

Received 0 Likes
on
0 Posts

Hi all,

I'd need some help with the following question:

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

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

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.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:

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

Join Date: May 1999

Location: Bristol, England

Age: 65

Posts: 1,808

Received 0 Likes
on
0 Posts

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)

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)

Join Date: Jan 2006

Location: Europe

Posts: 404

Likes: 0

Received 0 Likes
on
0 Posts

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.

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.

Join Date: Jan 2006

Location: Europe

Posts: 404

Likes: 0

Received 0 Likes
on
0 Posts

Hi there,

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

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!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

Join Date: Aug 2014

Location: Earth

Posts: 11

Likes: 0

Received 0 Likes
on
0 Posts

**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".

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".

Join Date: Feb 2002

Location: Sunny Solihull

Age: 67

Posts: 272

Likes: 0

Received 0 Likes
on
0 Posts

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.

Join Date: Jan 2006

Location: Europe

Posts: 404

Likes: 0

Received 0 Likes
on
0 Posts

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

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

Join Date: Jun 2002

Location: Wor Yerm

Age: 68

Posts: 4

Likes: 0

Received 0 Likes
on
0 Posts

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.

Join Date: Jan 2006

Location: Europe

Posts: 404

Likes: 0

Received 0 Likes
on
0 Posts

**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.Correct answer would be 510 kts.

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

Join Date: Feb 2002

Location: Sunny Solihull

Age: 67

Posts: 272

Likes: 0

Received 0 Likes
on
0 Posts

510 kts is correct.

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

The Machmeter only has instrument & pressure errors

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.