oh thats good and got one to go all the best...:ok:
I ve three written gnav,radios and met followed by two orals on gnav and radios.

GL Then ;) .

Oral on GNAV? Never heard of that.

You might get INS questions in Nav - and remember that the Nav exam has been traditionally a dustbin for all sorts of stuff that doesn't fit into other subjects - we tell our guys to expect the unexpected! :) Sometimes you might get a question that simply asks you to move figures around a formula.

One tip if you get a mention of an INS in Nav is to realise they are really talking about great circles.

Thank you so much Paco . Very helpfull info :D .
I will work on this stuff.

great info paco thanks a lot:)

Help ?
 

Can anyone please solve this question for me

Flying at FL270, an AWR weather return at range 40nm is identifiable when the centre of the beam is tilted between +1/-1deg & -6deg. What is the height of the base & top of the rain bearing cloud?

Are you sure that you are quoting the correct data?

This type of question usually includes a tilt up angle, a tilt down angle and a beam width. You can then use the range and the 1 in 60 rule to work out the solution.

But your quoted figures of between do not make any sense.

Hi, i'm a bit confused as on my CQB some questions of Air Law are out of date i suppose:

1) The VMC minima for an airspace classified as "B" above 10 000 feet MSL are: 8 KM CLEAR OF CLOUDS

Seems that now ICAO regulations are 8km 1.5 vis and 1000ft as according to jar ops..?

2)The protection areas associated with instrument approach procedures are determined with the assumption that turns are performed at a bank angle of:
SIGNED CORRECT:
25° or the bank angle giving a 3°/s turn rate, whichever is lower, for departure, approach or missed approach instrument procedures, as well as circling-to-land (with or without prescribed flight tracks).

What about this one??should be 20° for circling and 15° for missed app?

3)
According to JAR-FCL, Class 2 medical certificate for private pilots will be valid for SIGNED CORRECT:
60 months until age of 30, 24 months until age of 50, 12 months until age of 65 and 6 months thereafter

Shouldn't it be 60 until 40, 24 until 50 and 12 thereafter..?

4)
Runway edge lights excepted in the case of a displaced threshold shall be: SIGNED CORRECT:

Fixed variable white.

Isn't fixed variable white or yellow?

Thank you

The Oxford ATPL book of Principles of Flight has references that I need some help with.

For e.g. On Page 174; Chapter-Stalling
It talks about EASA 25.103(b) on a few occasions.

Now, where can I look up this regulation?

They are referring to the Certification Specifications for Large Aeroplanes CS-25. It would have been more appropriate to write "CS 25.103(b)". A copy can be downloaded from EASA's website, under Aviation Professionals//Legislation/Cetification Specifications if I remember correctly.

ICAO Annex III - Meteorological Service for International Air Navigation, Edition 15, Appendix V (Page 107)

TX25/13Z TN09/05Z
TX05/12Z TNM02/03Z

Maximum temperature 25°C at 1300Z Minimum temerature 09°C at 0500Z
Maximum temperature 05°C at 1200Z Minimum temerature -02°C at 0300Z

:)

Thanks hvogt. Found it.

ATP Nav Gen question help
 

Hi guys
I'm currently doing self-studiesfor the ATP preparation and need some helps from anyone who can give me hints to solve out this type of question:

An aircraft leaves position S01 20 W012 15 on a track of 360 degrees true for 01:06 mins.The heading is then altered to make good track of 090 degrees true for 03:12 mins to eestination.If the TAS is 320 kts and the wind is calm throughout the flight,What is the lat & long of the of the destination?

I just need the methodology to lead on how to solve the question

Many thanks in advance

NEXT TIME please post in the correct thread

HWB

320 kts for 1.1 hours is 352 nm. That's 352 minutes or 5 degrees 52 mins so you end up at the end of the Northern sector at N4 32, then turn right. The problem is now that 1 minute no longer represents 1 nm.......

3 hrs 12 mins at 320 kts is 1024 nm.

See if you can get the rest? Hint: you need to look at departure :)

From there its the Cos of Latitude time NM.

I take it that you are not self-studying for an EASA licence........because you can't!

The studying must be done through an Approved Training Organisation (ATO).

However, I've written out a painstaking, line by line solution below, for you to check your answer against.

As previously stated, 5°52' due north from S01°20' W012°15' lands you at N04°32' W012°15'.

To calculate the change of longitude from this position you need to use the "departure" formula:

Departure (NM) = Change of Longitude (mins) x Cosine of Latitude

The Departure (or distance) = 320 kts x 3.2 hours = 1,024 NM

1,024 NM = Change of Longitude (mins) x Cosine 04°32'

1,024 NM = Change of Longitude (mins) x 0.997 (Transpose formula)

Change of Longitude (mins) = 1,024 / 0.997

Change of Longitude (mins) = 1,027'

Change of Longitude = 1,027'/60' = 17.12°

Change of Longitude = 17°07'

To find out the Easterly longitude, subtract 012°15' from 17°07'

17°07' - 012°15' = E004°52'

Final Position = N04°32' E004°52'

Hope that helps. :)

Thanks very much guys i got the picture,it's so great

Hey guys,

Can anyone send me some examples with the answers regarding Drift Down, Im doing Performance second attempt and there were 2 questions on drift down but I havent seen any in the banks (using easaatp.com bristol and oxford). The driftdown questions were calculations, not just memorization.

I'm also finding different answers for calculating speeds from holding speeds and stall speeds, for prop and jet, any ideas?

Thanks

Given the following information determine the greatest distance from an obstacle, at the failure of one engine would cause an MRJT1 aircraft to clear the obstacle by the statutory minimum during the drift-down, the fuel used during the drift-down, and time taken to reach the obstacle.

Cruise Pressure Altitude 34,000 ft.
Gross Weight at Engine Failure 54,000 kg.
Ambient Temperature ISA +15°C.
Wind Component 40 kt tailwind.
Engine and Wing Anti-icing System on.
Air Conditioning System on throughout the drift-down.
Obstacle Pressure Altitude 14,000 ft.
One engine inoperative.

1. Select the appropriate graph for the given cruise altitude. For 34000 ft this is figure 4.25, which covers altitude between 33000 ft and 35000 ft.

2. Calculate the height required to clear the obstacle during the descent. The statutory minimum clearance during drift down is 2000 ft, so the required attitude at the obstacle is 14,000 ft + 2,000 ft = 16,000 ft.

3. Using the data in the table at the top left corner of the page adjust the gross weight at engine failure to reflect the conditions of the anti-icing and the air conditioning system. For the stated condition of engine and wing anti-icing systems on and air conditioning system on, the corrections are 54,000 kg + 5,650 kg (for anti-icing)= 59,650 kg. Note that an adjustment is required for air conditioning only when it is switched off, so no adjustment is required in this question.

4. Using the graph at the top right cornet of the page adjust the corrected gross weight at engine failure to reflect the temperature deviation. To do this enter the left edge of the sub-graph at 59,650 kg, then move parallel to the sloping lines to a point vertically above ISA +15°C. From this point move horizontally to the right edge of the sub-graph and read off equivalent gross weight at engine failure on the right vertical axis 61,500 kg.

5. Enter the right edge of the main graph and interpolate between 60000 kg and 65000 kg to locate the 61500 kg point. From this point draw a curve to the left interpolating between the printed curved lines on the graph. In doing this take care to take account of the fact that the curved lines converge as they move to the left.

6. Enter the left edge of the graph at 16,000 ft and draw a horizontal line to the left to intercept the curve drawn in stage 5 above.

7. At the intersection point interpolate between the dotted curved lines to estimate the fuel used during the drift-down. In this question the fuel used is approximately 1000 kg.

8. From the same intersection point draw a line vertically down to the bottom of the gridded area and read off the time required to drift-down. In this question it is approximately 29 minutes.

9. Extend the vertical line down to the reference line in the lower gridded area. From this point draw a line parallel to the sloping lines. From the existing 40 kt tailwind condition marked at the left edge of the graph, draw a horizontal line to the right to intercept the sloping line drawn previously. From the point of intersection of these lines drop vertically down to the bottom of the graph and read of the ground distance of approximately 175 nm.



If you are happy with the one above please try this one (something very like it appeared in a recent EASA ATPL exam)

Given the following information determine the greatest distance from an obstacle, at the failure of one engine would cause an MRJT1 aircraft to clear the obstacle by the statutory minimum during the drift-down, the fuel used during the drift-down, and time taken to reach the obstacle.

Cruise Pressure Altitude 37,000 ft.
Gross Weight at Engine Failure 44,000 kg.
Ambient Temperature ISA -15°C.
Wind Component 30 kt headwind.
Engine Anti-icing System on.
Wing anti-icing off.
Air Conditioning System on throughout the drift-down.
Obstacle Pressure Altitude 23,000 ft.
One engine inoperative.

Answers:

Ground distance 122 nm.
Time = 24 minutes.
Fuel used = 700 kg.

Here's a question few days ago i found out the answer but forgot the methodology that led me to this answer:

The scale at the equator on a Mercator chart is 1cm to 5nm.How many nm to the inch are there at S 44 00?

Start by converting 5 nm/cm at the Equator into nm/inch

There are approximately 2.54 cm in an inch so multiplying 5 nm by 2.54 gives 12.7 nm/inch at the Equator.


Then use the following equation to calculate the distance per inch at 44 degree latitude.

Distance at B = ( Distance at A x Cos B) / Cos A

Where A and B are latitudes.

Using A = 0 latitude at the Equator and B = 44 degrees latitude we have

Distance at 44 degrees = (12.7 nm x Cos 44) / Cos 0

Distance at 44 degrees = 9.14 nm

So the scale at 44 degrees is 9.14 nm/inch

Thank God I passed my UK CAA ATPL, many years ago
 

I would never pass muster these days....

I seriously wonder just what practical benefit these questions actually give to the human race.

I am impressed that my replacement in a few years time will actually have passed these hurdles.

Good luck guys and gals.

Haha. Absolutely true. Well there are more questions that are totally useless.

I was actually wondering if there are any pilots who use Grid Navigation in there daily life as a pilot?

Sunrise / Sunset times. Stuff like that.

I am also very happy that I passed all exams first Attempt few weeks ago. Nav. With score 94%.

Had also some of these converting questions of scale. And actually it is not very hard, but you have to see how it works. Since I am from the Netherlands, I always converted it to Metric, and from there find out the scale. Scored 100% on that part.

You'd be surprised. I'm using convergency/change of longitude all the time in my head.

Sunrise/sunset, too, although I don't use the almanac, just the GPS

Well Convergency Conversion I can imagine, but Grid?

Yeah I gues no pilot will take Almanac with him or her ;) .

Hi

Q. For which of the following is a flight plan, in accordance with Annex 2, to be submitted?

a) Any flight crossing an IFR boundary
b) Any IFR flight in class F airspace
c) Any flight in controlled airspace
d) Any flight more than 40 km from the coast

b) marked correct but whats wrong with (a) or (c).

Thanks

According to 3.3.1.2 c) of ICAO Annex 2, a flight plan is required for "any flight across international borders", however there is no such thing as an "IFR boundary"; hence, answer a) is wrong.

Answer c) is wrong because a flight plan must be filed for "any flight or portion thereof to be provided with air traffic control service" (3.3.1.2 a) of ICAO Annex 2), not for any flight within controlled airspace. Think of VFR flights in airspace class E - controlled airspace, but no clearance or flight plan required.

Thanks hvogt

May I also ask you:

In an advisory airspace, IFR traffic using the advisory service will be separated from other participating IFR traffic using the advisory service.

Is this universal or specific to Europe?

All the references I've seen just say that separation provided is between IFR flights and don't mention whether they are participating or not.

Thanks

Haroon

The term "separation" is somewhat relative when used with air traffic advisory service because this service will not issue clearances, it will only give advice and suggestions. Needless to say, separation based on such suggestions can only be achieved between aircraft which are known to the unit providing them. In other words, aircraft which are not "participating" cannot be separated. Keep in mind that ICAO permits IFR flights in airspace class F without clearance, so in airspace class F there might be a wild mix of IFR and VFR flights known or unknown to the air traffic services unit in charge.

As for your question concerning universal or European application of the respective rules, each state will have its own rules, some do not even have class F airspace (Denmark and Chile if I'm not mistaken), others might not provide air traffic advisory service at all but offer flight information service instead. Here in Germany, for example, IFR flights in airspace class F are subject to an ATC clearance and there will never be more than one IFR flight at the same time in the same airspace.

Thanks hvogt, much appreciated

If there is sufficient time for information to be disseminated by other means, a NOTAM is not issued. What is the time limit?

2 days
28 days
14 days
7 days Marked Correct

i cant find any reference for this, can someone help pls

thanks

Hi standing by for the answer regarding the previous question (post #374).

Got some more:

1) What documentation is required by persons travelling by air, for entry into a state?

Passport and visa
Passport and confirmation of inclusion on the general declaration passenger manifest
The same as would be required if the person arrived by ship
Passport, visa and any necessary health documentation (vaccination certificates)

Why not the first option or the last option?


2) Who is responsible for setting into movement the Alert Phase?

ATC and the FIR
The State and ATC
The Area Control and the RCC
RCC and the FIR

Do they mean FIC? Why not the first option?


:arrow: Which of the following applies to general declaration?

Accepted orally for crew or passenger baggage
Accepted orally for crew baggage
Accepted orally for passenger baggage
Never accepted orally

Cant find a reference to this one!

thanks

That about visa Passport.
Its a useless question.Like most questions in AL and OPS. Just remember.

Have also question about, what is MIN age of a Cabin Crew? So as you can see. These kind of questions you should just remember. There is no logic why you should know that.
Questions about seperation , those are usefull , and signs, colours etc.

need some logic to remember otherwise will forget when the time comes :)

No logic needed in AL and OPS.

Other subjects, yes then you should definetely understand that theory, So not only QB but also study your books. Otherwise its hard to get through the exams. But AL and OPS are very dry stuff.

Hi guys
I've got this one here:

Flight time 03:30
Reserve fuel shouldn't be less than 25% of the trip fuel
Block fuel:260 kgs
Taxi. :10 kgs

How many kgs of fuel should remain after 2 hours?
Any help is appreciated,i'm looking for an easy method to solve this kind

Take-off fuel is block fuel (260 kg) minus taxi fuel (10 kg), i.e. 250 kg.

Reserve fuel must not be less than 25 % of the trip fuel, so reserve fuel is (250 kg / 125) · 25 = 50 kg.

Trip fuel is take-off fuel minus reserve fuel, i.e. 250 kg - 50 kg = 200 kg.

After 2 hours of a 3.5 hours’ flight, 2/3.5 of the trip fuel are burnt: 2/3.5 · 200 kg ≈ 114 kg. The remaining trip fuel is hence 200 kg - 114 kg = 86 kg. The 50 kg reserve fuel must not be forgotten, so remaining fuel in tanks is 86 kg + 50 kg = 136 kg.

The simple way, least calculator hit (other methods are available) I teach my students is:-

Take-off fuel / Total flight time * Time remaining.

250 / 3.5 * 1.5 = 107 kg.

The question usually says 25% reserve of REMAINING trip fuel, which means the actual amount of reserve fuel required is decreasing as the flight progresses and would be zero on arrival.

As this question didn't ask for a split of trip & reserve the answer is 107 kg (assuming it meant REMAINING). However to split 107 / 1.25 (back work 25%) = 86 kg trip and reserve would be 21 kg which is 25% of 86 kg.

This is a proportion question and the old whiz wheel excels at this problem.

RichardH, I don`t absolutely agree with you. Trip fuel is fuel for all the trip, which is 3:30. If the final reserve should be 25% Trip, it is related to the Total Trip fuel, not just remaining time, as you say.
Your theory would say, that after landing, you can have zero final reserve, which you surely know is not true.
So after my calculations, the result is also 136.
Please, anyone correct me if I`m wrong.

This type of question has always caused issues like this as it is a fuel related question but not fuel policy directly.

It's really a light aircraft fuel question where a reserve of REMAINING trip is perfectly okay, though strange to get your head around especially landing in THEORY with no reserve, but of course you started with that 25% and assuming nothing happened then that reserve would be intact on landing.

Questions similar to this have never mentioned FINAL RESERVE as I should agree with you if it did. Trust me with years of instructing experience and knowing what the examiners are after this is a proportion question not fuel policy knowledge, though perhaps not the best way to test it by ignoring the "rules". The examiners are basically saying with inaccurate fuel gauges the reserve is set at 25%, you don't know how that is split.