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 :)