ATPL theory questions

[RedBullGaveMeWings]

I am mainly using Aviation Exam on Exhaustive mode. I will also start integrating with Bristol QB, which is very good if you are sitting your exams under the UK CAA.

[RedBullGaveMeWings]

Binary code is no mistery to me and I have some experience in programming, but curiosity strikes me now... Why is EASA asking to convert a binary code?

[paco]

It's a throwback from the days when you were expected to fix your own radios in the jungle......

[KayPam]

Hello

I just started performance
Could someone please confirm that the entire CAP698 document will be available during the exam ?
My book says things like : "this is included in CAP698 so it does not need to be learnt"

Thanks

[Alex Whittingham]

That would not be so in the UK. The UK CAA used to allow you to take the whole CAP in but this has not been the case for some while, now you only get an 'Annex' showing the relevant graph. Our students, therefore, do need to learn the various formulae and conversion factors. If you are sitting the exam in France you should ask the DGAC what their rules are, or ask your ATO.

[RedBullGaveMeWings]

I am a bit confused about the following question:

For a given IAS and angle of aileron deflection, increasing altitude will:

A) increase the rate of roll [correct answer]
B) increase the rate of turn
C) reduce the rate of turn
D) reduce the rate of roll

Clearly A is correct because aerodynamic damping is less at high altitude, however, constant IAS means an increasing TAS which also reduces rate of turn.

What does examiner really want to ask here? I see two correct answers.

[Alex Whittingham]

Not being an aerodynamicist I'd have gone for (D). I remember exactly this exercise in the Hawk when an instructor tried to convince me that the highest rate of roll occurred at altitude, when it clearly didn't. On landing I found the reference in the manual which said the optimum rate of roll was at sea level. I always assumed he had written the lesson down incorrectly on his knee pad, but who knows.

The clue might be in the 'constant IAS' phrase, you get better roll rates at high IAS at sea level but you can't make those high IAS's at height because of Mach effects. I will wait for an aerodynamics expert...

[KayPam]

Alex : thanks

RBGMW : You might want to derive a simplified flight mechanics formula
The rolling moment equation is (simplified)
Inertia*φ'' =(Cldl*dl + Clp pb/2v) * QSb

Any answer to a rolling question is behind this equation.
Then you just have to interpret your question.
Are we talking about the maximum achievable rate of roll ?
If yes, then phi''=0 when max rate of roll is reached and :
Cldl * dlmax = Clp pmax*b/2V

Cldl, dlmax, Clp, b are constant.
So pmax will increase if V will increase. At constant IAS, V will increase with altitude

The rate of turn is unrelated to the angle of aileron deflection.
Depending on the aircraft, you will be in a steady turn with aileron undeflected or deflected in any direction (towards the turn, outwards the turn, more directions if they existed)

So I'd go A

[Alex Whittingham]

Thanks KayPam, that in a nutshell, is why I don't teach principles of flight! will work through it slowly tomorrow.

[keith williams]

This question is intended to be answered by ATPL candidates, it should be possible to solve it without going beyond the ATPL syllabus.

The question is an old one (probably 16 or 17 years old), so it really should be bread and butter stuff for ATPL instructors and students.

Let’s start by looking at a few things which ATPL students should know:

1. Rolling motions are generated by the aerodynamic forces acting on deflected ailerons.

2. The magnitude of these aerodynamic forces increases with increasing angle of deflection and with
increasing dynamic pressure.

3. If we climb at constant IAS the dynamic pressure will remain approximately constant, so the rolling
forces generated by any given aileron deflection angle should also be approximately constant.

4. Rolling motions are opposed by aerodynamic damping.

5. Aerodynamic damping decreases as TAS increases.

6. Climbing at constant IAS causes the TAS to increase.

7. Radius of turn = TAS squared / g Tan AOB.

8. Rate of Turn = g Tan AOB / TAS.

9. Under normal circumstances, keeping the ailerons deflected will cause continuous
rolling motion, but will not produce a constant angle of bank.

Looking at statements 1, 2, 3, 4 5 and 6, we should see that climbing at constant IAS will cause roll rates at any given aileron deflection to increase (Option A).

Looking at statements 7, 8 and 9 we should see that aileron deflection does not control the rate of turn or radius of turn, so options B and C are incorrect.

At this point we may be tempted to argue that “if we fly high enough our maximum IAS will be reduced, so our maximum roll rate will be reduced. But the question specified “ constant IAS” so this excludes any altitude at which it is not possible to maintain a constant IAS. So option D is incorrect.

It is a well known fact that many of the questions used in these exams are defective, but candidates must take care to avoid being too eager to simply tell themselves “Oh this is just another duff question”. Such eagerness is likely to cause them to raise large numbers of appeals, which are subsequently rejected. If they have paid insufficient attention to selecting the best option, the result is likely to be an examination failure. Whenever a candidate thinks that two or more options are correct, their first action should be to look very carefully at why they think each option is correct.

[Alex Whittingham]

Thank you Keith. I suspect that the clue to my misunderstanding is indeed the IAS thing for when comparing roll rates at 420KT IAS/TAS at sea level with roll rates at 200KT IAS/420KT TAS at height the sea level roll rate would win out. Hence the statement in the flight manual that for a particular aircraft (albeit a reasonably fast little jet) the optimum rate of roll is at high IAS at sea level.

[oggers]

I agree that the correct answer is A.

Given that IAS and aileron deflection are constant it follows that:

The dynamic pressure is constant and therefore roll moment due to aileron deflection is constant.

Given that altitude increased (with IAS constant) it follows that:

TAS increased.

Steady state roll rate is achieved when roll moment is in equilibium with roll damping. And for a given roll rate, roll damping would be less at higher TAS. Therefore roll will go to a higher rate before equilibrium is reached.

As RedBull says, answer C is also tempting because the higher TAS would result in a reduced rate of turn as well. But the question does not specify any angle of bank or even that the aircraft is turning. It could be an aileron roll. So based on the information available in the question there is only one answer that is definitely correct, and that is A.

[RedBullGaveMeWings]

Now it's crystal clear, thanks Keith.

[KayPam]

Could someone please explain ?

https://i.gyazo.com/76c101cae12e601f9756ca777c79108f.png

Obviously, V1 can't be lower than Vmcg.
But who says one has to choose a balanced V1 ?

In this case, one would just increase V1 above the balanced value. As a result, the ASD and ASDR would increase above the one engine out takeoff distance.

Anyone has an explanation in favor of the green answer ?

[keith williams]

This question has been around for many years, and I recall appealing it as long ago as 2007.

The UK CAA examiners' answer at that time was "Ah but you must assume that the aircraft is taking-off at it's field limited take-off mass, so there is no space available to increase ASDR" When I pointed out that nothing in the question indicated it being a field limited take-off, there was no convincing reply. I believe that most of the schools appealed the same question and got the same response. This was quite common at the time.

[KayPam]

Thanks Keith !

I will be passing (not taking :p) instruments, meteorology, performance and mass and balance next week.
My books were read and all AvExam questions are answered, so my method requirements are met

However, for the next subjects to come, I'm not so sure if I should still apply the same method...
Next set of exams will be flight planning, principles of flight, and aircraft general knowledge.
For AGK, i'm pretty sure it's best to read the entire books and to do all the questions.
But flight planning seems really easy - just time consuming
And POF is sort of in the middle I think.*

I am doing the flight planning questions right now, without having ever opened the books, and I find myself answering correctly to 90-100% of the available questions..
Overall there are just the same calculations to do over and over again with few variations and little specific knowledge to have (like what are the different fuel quantities required, how long before a flight should one file a flight plan and that's pretty much all..), and then careful thinking and calculations will guarantee the correct answer..

I calculated I would be spending like 15 hours just answering AviationExam questions on this topic...

Is it really necessary to see all avexam questions ? Could there be any surprise question like there could be in a subject like AGK, instruments or meteorology ?

Fellow ATPL students, does your method include answering all the questions on the question bank ?


*Note that I'm currently working for airbus flight testing department so I'm well used to calculations and the sort of tables/graph they provide in this 33-test.

Following subjects will be human factor, gen nav and radio nav.
Will it be beneficial to have seen all the questions on these subjects ? I believe it could very well be so.

By the way, here is an advice for all distance and heading calculations of flight planning :
Instead of looking up the points and measuring your map distance then measure it against a meridian to have ground distances, like a galley slave would do, simply use the approximate or exact formula for distance computation based on the coordinates (which are given in the question) :
Pythagore (assumes earth is flat, which works very well for distances under 1000-2000 km) : 1 degree of latitude is 60nm and 1 degree of longitude is 60*cos(latitude)nm
Or, for very long distances :
D=Rt*acos(cos(lat1)cos(lat2)*cos(lon1-lon2)+sin(lon1)*sin(lon2))
Where Rt is your earth radius in the chosen unit.
(This formula is very easy to remember, just remember lat lon then 1 2, 1 2, 1 2, works just as well with 2 1, 2 1, 2 1)
And simple trigonometry (in the plane!) to get the true direction.

[paco]

Calculations are often easier than measurements. You can do such triangles on the front face of the Jeppesen CR-3.

And with the ED-6, remember that it is a 515000 not a 500000, so use the scale marker on the map if you do measure things.

[GcitoG]

Hi y'all

I have a job interview next week and have been practising with Prepware FAA ATP 2017 question bank. Chances are the airline has not update theirs yet, so my question goes like this: "Are there many changes on the new bank so that I can know what to expect?"

[oggers]

KP

Anyone has an explanation in favor of the green answer ?

Yes:

"If the balanced value of V1 is found to be lower than Vmcg, the take-off is not permitted"

...is a correct statement.

[KayPam]

Sorry but I don't see the point.

V1 must be higher than Vmcg is a correct statement.

If the balanced value of V1 is lower than Vmcg, just increase V1 up to Vmcg !!
Then if the ASD is still within the limits of the runway (including required margins) then you can takeoff !

[oggers]

KP

Sorry but I don't see the point.

V1 must be higher than Vmcg is a correct statement.

"V1 must be higher than Vmcg" is not one of the 4 answers you have to choose from. "The take-off is not permitted" is.

[Alex Whittingham]

I think Kay Pam's point is that, if the field is 'artificially' balanced by assuming that ASDA = TODA and there is extra ASDA available, then a solution would be to discard the 'balanced' solution and work out a FLL TOM and V1 for the real field lengths. As the effect of increasing ASDA is to increase both the FLL TOM and V1 it may be possible to either (i) use the new V1 at a higher TOM without modification if it is naturally greater than VMCG or (ii) choose a high V1 within the acceptable range so that it is equal or greater than VMCG if the aircraft is intended to operate at the 'balanced' FLL TOM. If either of those were possible then A becomes the correct answer. As Keith says there is not enough information in the question to say whether there is additional ASDA or not, or what the intended TOM might be.

[keith williams]

And if the aircraft is not field limited, climb limited nor obstacle limited we could do a reduced thrust take-off, which might also make options B and C correct.

[Alex Whittingham]

Ah, not sure about C there Keith, but certainly B if a derate is used rather than flex. But who knows?

[KayPam]

And if the aircraft is not field limited, climb limited nor obstacle limited we could do a reduced thrust take-off, which might also make options B and C correct.

B is certain, C depends on the wording I guess (ASDR before or after the new assumption of reducing thrust)
You sir are entirely right, all answers could be correct, depending on the situation.

Where is the LOL emoticon when you need it ?

[oggers]

Yes Alex I understand take-off performance. My point is that it doesn't matter what alternative answers "may be possible" because one of the answers is correct anyway, based solely on the limited information provided.

KP asked "if anyone has an explanation in favour of the green answer". There is one and it is the only one that does not involve changing the question from "which of the following is correct" to "which of the following would be correct if V1 was no longer found to be less than Vmcg".

Maybe after studying performance people feel insulted by such a simple question that they have to change it around so as to try and show off. It seems to me that this question comes straight out of CAP698, where it says:

Compare V1 with VMCG. If V1 is less than VMCG, take-off is not permitted

My experience with these multi choice tests is that the questions are often based on literally regurgitating simple statements taken from approved study material. Sure, the answer may be less complete or less clever than you would give in an oral or practical test, but what matters in trying to pass the multi choice is simply to choose the answer that will get you the mark.

I don't wish to mire the atpl questions thread in a subjective debate about the merit of this question. But if anyone wishes to raise a thread about it in the Technical or Questions forums there is probably a lot more that could be said.

[KayPam]

"Maybe after studying performance people feel insulted by such a simple question that they have to change it around so as to try and show off. It seems to me that this question comes straight out of CAP698, where it says:

Compare V1 with VMCG. If V1 is less than VMCG, take-off is not permitted"

This is the best explanation to this particular choice of correct answer thanks.

[KayPam]

I will be passing (not taking :p) instruments, meteorology, performance and mass and balance next week.
My books were read and all AvExam questions are answered, so my method requirements are met

However, for the next subjects to come, I'm not so sure if I should still apply the same method...
Next set of exams will be flight planning, principles of flight, and aircraft general knowledge.
For AGK, i'm pretty sure it's best to read the entire books and to do all the questions.
But flight planning seems really easy - just time consuming
And POF is sort of in the middle I think.*

I am doing the flight planning questions right now, without having ever opened the books, and I find myself answering correctly to 90-100% of the available questions..
Overall there are just the same calculations to do over and over again with few variations and little specific knowledge to have (like what are the different fuel quantities required, how long before a flight should one file a flight plan and that's pretty much all..), and then careful thinking and calculations will guarantee the correct answer..

I calculated I would be spending like 15 hours just answering AviationExam questions on this topic...

Is it really necessary to see all avexam questions ? Could there be any surprise question like there could be in a subject like AGK, instruments or meteorology ?

Fellow ATPL students, does your method include answering all the questions on the question bank ?


Following subjects will be human factor, gen nav and radio nav.
Will it be beneficial to have seen all the questions on these subjects ? I believe it could very well be so.

Hello,
I passed 3 out of 4 subjects today.

Please be reassured, I haven't failed the fourth one, but there was no time to take it today, tomorrow then (as planned). Mass and balance so no worries.

The above questions are hot in my head right now.

Thanks

[RedBullGaveMeWings]

I must be missing something here or... :confused:

One of the following statements about aircraft ground movement is correct:

A - a taxiing aircraft has priority over a vehicle towing an aircraft (marked correct)
B - a vehicle towing an aircraft has priority over a taxiing aircraft (my answer)
C - An aircraft overtaking another does so by passing on the right
D - Two airplanes approaching head-on will alter course to the left

[oggers]

'A' is correct according to SERA.

[RedBullGaveMeWings]

To which SERA document should I refer to?

[oggers]

Commision Implementing Regulation No 923/2012; SERA.3210 Right-of-way.

(d).(4).(iv).(A) vehicles and vehicles towing aircraft shall give way to aircraft which are landing, taking off, taxiing or being towed;

So a taxiing aircraft has priority over a vehicle towing an aircraft.

As an aside, it also says in effect that vehicles towing aircraft shall give way to aircraft being towed by vehicles. Now that really is :confused:

[RedBullGaveMeWings]

Ok I found it. That question omitted to be referring to an EASA regulation instead of ICAO Annex 2.

[G DROP]

Has anyone come across the Met question "how many hurricanes are there east of Darwin?" and any idea how to answer it?

[KayPam]

Learn by heart the answer..
It will fly out as soon as the test is passed.

Or you could get an australian friend and call them regularly and see how many times per year they're really worried about their wooden house.

[G DROP]

I can't find it on the banks. I've heard CAA are asking it in recent exams though - no one seems to be sure of the answer

[PPRuNe Towers]

The hurricane question.

If it is Darwin, Australia not sure there will ever have been one. Cyclone perhaps rather than a Hurricane.

Rob

[paco]

There aren't any - in Australia, they are called Severe Tropical Cyclones according to their government's Met book. Did the question include a timescale? They occur between November to April.

Of course, it depends how far East you go - if you carry on flying you will meet hurricanes off the Eastern seaboard of the US :)

[G DROP]

Sorry English isn't my first language so my mistake for the confusion. I have friends in an integrated course sitting the MET exam soon - I'll see what I can find out!!

[talk2jamesbond]

Hi Dear all,
I have some ATPL questions, If anyone could help me would be very appreciated.

Q.1 You are flying at FL 130, and your true altitude is 12000 ft. What is the temperature deviation from that of the standard atmosphere at FL 130 (QNH 1013.2 HPA) ?

(a) ISA -20 C
(b) ISA +/- 0 C
(c) ISA +20 C
(d) ISA +12 C

If you could give me, how do you calculate would be appreciated.

Q.2 TAF LSZH 250716 00000KT 0100 FG VV001 BECMG 0810 0800 VV002
BECMG 1012 23005KT 2500 BKN TEMPO 1316 6000 SCT 007= Which of these statement
best describes the weather that can be expected at 1200 UTC

(a) Meteorological visibility 6 kilometers, cloud base 500 feet, wind speed 5 knots
(b) Meteorological visibility 2.5 kilometers, cloud base 500 feet, wind speed 5 knots
(c) Meteorological visibility 800 meters, wind from 230, could base 500 feet
(d) Meteorological visibility 800 meters, vertical visibility 200 feet, calm

I think the answer is (b) however answer is (a), Maybe am wrong, could some one explain ?

Thank you very much everyone :)

[RedBullGaveMeWings]

In the first question, you have not been given any other barometric values, so it has to be A. It must be colder than ISA for your altimeter to overread. 4% for every 10 degrees of ISA deviation fits the ISA -20° C. No calculations are needed with a bit of exam technique.

The TAF question, I would say it's B. Where did you find these questions?

[talk2jamesbond]

Dear RedBullGaveMeWings

This is EASA ATPL questions :)

[RedBullGaveMeWings]

Yeah I can see that, I actually meant to ask in which question bank you found them.

[oggers]

I think the answer is (b) however answer is (a), Maybe am wrong, could some one explain ?

Q2; you are correct it must be b. It cannot be a. However there is no such group as BKN without height eg BKN005.

Suggest you find a better question bank!

[keith williams]

This question has been around for many years.

The original version included BKN005 and the correct answer is option B.

[felixflyer]

What happens after leaving feedback on an exam question.

I had a question in Air Law asking what needs to be taken into consideration for an IFR flight. The options were all applicable so I chose one and left feedback.

I am not sure whether I got this correct or wrong. Do I get feedback on feedback?

[mike172]

Were the options along the lines of weather, fuel, flight plan...?

I had the same question in flight planning when I sat my exams. Left feedback but my mark never changed so either I guessed correctly or they didn't read my feedback. Not sure how it works but I always said to other candidates always leave feedback on ambiguous or poorly written questions. The thinking being that the more who leave feedback the better the chances of a question being reviewed. Or so you would think...

I did have marks in two exams go up by a few percent so it does happen.

Some of the questions are a joke, though, and candidates shouldn't be having to flag poor questions up for review. Especially when you're forking out close to 70 quid just the sit the bloody things.

[ersa]

Keeflyer is spot on, unless someone pays for a review , nothing happens, leave as many comments as you like nothing will happen

[felixflyer]

Yes, that was the question.

I got a decent mark so it's not an issue, I was just curious what happens.

Sounds like its a bit of a waste of time.

[RedBullGaveMeWings]

Post deleted. I have just realised where I was mistaken.

[RedBullGaveMeWings]

OK now this one is weird...

Found on Aviation Exam, ID 26537

Given

Mach 0.80
Flight Level FL330
OAT: ISA+15

TAS is approximately (compressibility factor of 0.94):

A 265 kts
B 480 kts [this is what I get with formulas and CR-3)
C 420 kts
D 450 kts (marked correct)

Is it me or EASA?:ooh:
Of course if I multiply 480 by 0.94 I get a approximately 450 kts... but it is not correct, is it?

[Thegreenmachine]

Not an expert but here goes...
I got a temperature of -36degrees Celsius which is 237 kelvin
38.95 x square root237
X . 8
X .95
= 450.91

[KayPam]

About aviation exam, 33 (flight planning)
Could anyone confirm that many questions (even types of question) are completely absent ?
For example, I saw no question about which route to choose between two waypoints, or what the distance and heading are between two waypoints of given coordinates.
Whereas I saw these questions in my school's final test.

Is it normal ?

[paco]

The temperature would be -66*, so if you set the Mach index arrow against that and look against 0.8 on the inner scale of the whizzie you should see 446 kts or so on the outer scale.

*33 x 2 = 66 -15 +15

[Alex Whittingham]

er, temperature is (+15) - 66 +15 = -36. I agree with KayPam, why would you apply compressibility?

[RedBullGaveMeWings]

The temperature would be -66*, so if you set the Mach index arrow against that and look against 0.8 on the inner scale of the whizzie you should see 446 kts or so on the outer scale.

*33 x 2 = 66 -15 +15
It's ISA+15, not -15.

[paco]

33000 feet at 2 per thousand feet = 66.

Starting at +15 on the surface means the temp should be -51, which is logical, given that it's supposed to be -56.5 at 36 090 feet.

But it's ISA +15 so add it back......

Looks like one of those questions where they include false information.

[RedBullGaveMeWings]

Err, doesn't ISA+15 mean that it is warmer by 15 degrees? I keep getting OAT -36° C which are 237° K.

LSS=38.95xsqrt237=approximately 600 kts
600x0.8=approximately 480 approximately.

Same with CR-3. It takes compressibility into account, doesn't it?

[paco]

Duh, of course.... early morning is my excuse! :)

-36 it is, the CR3 (or CRP5) Mach index should take that into account. And you're right that makes nearly 480, using the Mach Index. But then, applying 0.94 takes you back to roughly 450, which I believe would be the wrong answer because they didn't include CAS in the stem. But who knows?

[Alex Whittingham]

I think the question KayPam is asking is, why would you apply compressibilty? That correction comes between CAS and EAS, not TAS and Mach. Having found your TAS you would apply compressibility if you were working backwards to find CAS.

[KayPam]

Regarding this 480kt vs 450kt wierdness:
https://i.gyazo.com/0401a1f0cda2db361a379b7f0ffe4008.png

I put the question's numbers in the system, and TAS is indeed given as 480kt.
And, if you compute EAS/CAS, you will find 0.94.

So, the best I can come up with to explain the whole situation is that they made a mistake in thinking the compressibility correction should be applied to the TAS whereas it should be applied to the CAS only.

A situation where the correct answer would be 450kt is the following :
Mach number is measured from (Pt-Ps)/Ps.
So, say you don't have any device to correct your machmeter for compressibility (which is a problem in itself.. but let's assume that).
The machmeter will overread, due to compressibility.

Let's see how it will overread in details :
You are flying at 450 KTAS (but don't know it yet)
You measure Pt, and find you're flying at 265 KCAS (and 252KEAS in reality)
However, since you're measuring a Pt that's higher than the real one, you're actually not accounting for compressibility, and whereas your EAS is 252kt, you mistakenly believe you have an EAS of 265kt
When you correct for density to get TAS, you compute TAS from an EAS that's higher than the real one, so TAS is higher than your real TAS, close to 480kt.
Then you compute your mach number and find a number close to 0.8.
So you believe you're flying at M0.8

Your machmeter measures M0.8, but it actually overread due to non accounting of compressibility.
At this precise moment, you're in your imaginary aircraft with your wierdo-non-corrected-machmeter and an EASA civil servant steps into the cockpit and asserts the compressibility factor is 0.94. Darn, your non-corrected-machmeter overestimated speed by a factor of 6% !
So you're flying at M.75, which is exactly 450kt.

I reckon it is very very far fetched and I don't know what I would answer if this question did come up.
I don't even know if a non corrected machmeter exists...
Aviation Calculator (http://www.hochwarth.com/misc/AviationCalculator.html)

[paco]

It is a mistake because if you use the mach index, you are reading TAS directly off the flight computer, and the compressibilty calculation is to give you TAS anyway.

[RedBullGaveMeWings]

Thanks KayPam... and thanks EASA:(

[KayPam]

It is a mistake because if you use the mach index, you are reading TAS directly off the flight computer, and the compressibilty calculation is to give you TAS anyway.

It isn't the first surprise EASA offers :ugh:

[selfin]

KayPam,

And, if you compute EAS/CAS, you will find 0.94.

There is a similar question reported here (http://www.pprune.org/professional-pilot-training-includes-ground-studies/455580-atpl-theory-questions-27.html#post8619922) and, as noted by numerous responses immediately above, the factor is completely irrelevant to the question, and the answer marked correct above is clearly wrong. Nevertheless, the figure is equivalent to the quotient of two compressibility factors f/f_0 using equations 11, 16, and 20, in NACA Report 837 (http://naca.central.cranfield.ac.uk/report.php?NID=2451) (Aiken, W., 1946). The ratio depends only on pressure altitude and Mach number and is therefore independent of temperature deviation.

Mach number is measured from (Pt-Ps)/Ps.
So, say you don't have any device to correct your machmeter for compressibility (which is a problem in itself.. but let's assume that).
The machmeter will overread, due to compressibility.

There is no Mach number in an incompressible flow because the speed of sound is infinite. Neither Mach meters nor airspeed indicators need to be "corrected for compressibility" so it does not necessarily follow that determining Mach number by a quantity such as `(Pt - Ps) / Ps` will result in an error. The Mach number can be related to the normalised pressure rise `(Pt - Ps) / Ps` using a simple algebraic expression without needing to adopt an incompressible flow assumption. The applicable technical standard for Mach meters (SAE AS8018 (http://standards.sae.org/as8018a/) citing NASA Technical Note D-822 (https://www.mikrocontroller.net/attachment/115180/NASA_Technical_Note_D-822.pdf)) assumes an isentropic flow in relating pressure rise to Mach number. The same is true for airspeed indicators, although in their case the impact pressure q_c–the pressure rise–is normalised by standard sea-level pressure. Impact pressure (for M ≤ 1) is related to Mach number by:

M^2 = 5*(-1+(1+q_c/p)^(2/7)),

and to true airspeed v by:

v^2 = 7*(p/rho)*(-1+(1+q_c/p)^(2/7)), rho is ambient air density.

Because these equations are based on an isentropic model they account for variation in density as air is brought to rest in a pitot probe. The latter was first given in this form by Saint-Venant and Wantzel in Journal de l'École polytechnique, vol 16, 1839 (http://gallica.bnf.fr/ark:/12148/bpt6k4336833/f86.item) (according to Stanton [link] (http://rspa.royalsocietypublishing.org/content/111/758/306)).

In an incompressible flow the relationship for true airspeed squared is simply,

v^2 = 2*q/rho,

and q is dynamic pressure which is the difference between total and static pressures in an incompressible flow. This incompressible flow model is not used by Mach meters or airspeed indicators. Dynamic pressure may be thought of as a first order approximation of impact pressure - see for example the paragraph following eqn 2, and particularly figure 2, in NACA Report 247 (http://naca.central.cranfield.ac.uk/reports/1927/naca-report-247.pdf) (Zahm, A., 1926). The inverse relationships for impact and dynamic pressures are,

Isentropic flow: q_c = p*(-1+(1+(1/7)*(rho/p)*v^2)^(7/2)), and,
Incompressible flow: q = (1/2)*rho*v^2.

The Maclaurin expansion of p*(-1+(1+(1/7)*(rho/p)*v^2)^(7/2)) about v^2 = 0 is (1/2)*rho*v^2 which is the Euler-Bernoulli solution to Euler's equation for an incompressible flow.

The first use of a compressibility factor, of which I'm aware, in a circular slide rule for obtaining TAS from CAS, as in eqn 9 in NASA Technical Note D-822 or eqn 20 in NACA Report 837, is described in Ezra Kotcher's (of Bell X-1 fame) US patent 2342674 (https://docs.google.com/viewer?url=patentimages.storage.googleapis.com/pdfs/US2342674.pdf) of 29th Feb 1944 for an Air Speed Computer. For on the influence of errors see NACA Technical Note 1605 (https://ntrs.nasa.gov/search.jsp?R=19930090948) (Huston, W., 1948), Accuracy of Airspeed Measurements and Flight Calibration Procedures.

[paco]

I've lost the will to live already..... :)

I've checked with a couple of other senior instructors, and they would expect the answer to be 480 knots if it was encountered in an Instruments exam.

[Alex Whittingham]

I, for one, am fascinated as selfin's explanation goes some way to explaining something that has worried me for some time. I have always found it difficult to reconcile the two statements:

1. On aircraft with Air Data Computers the ASI displays CAS
2. The Air Data Computer calculates CAS using variations of Saint Venant's formulae which assume compressible flow.

Because if #2 were true, the ADC would output what I know as EAS, CAS corrected for compressibility, although it still seems to be called CAS. Is this the case, selfin?

[KayPam]

There is no Mach number in an incompressible flow because the speed of sound is infinite. Neither Mach meters nor airspeed indicators need to be "corrected for compressibility" so it does not necessarily follow that determining Mach number by a quantity such as `(Pt - Ps) / Ps` will result in an error. The Mach number can be related to the normalised pressure rise `(Pt - Ps) / Ps` using a simple algebraic expression without needing to adopt an incompressible flow assumption. The applicable technical standard for Mach meters
Nothing prevents you from calculating a mach number, from speed and temperature, even in an incompressible flow.
Nothing would prevent you from building a machmeter which would would exactly compute the mach from the following formula :

TAS=sqrt(2(Pt-Ps)/rho)
LSS=sqrt(gamma R T)
Rho=Ps/RT
TAS/LSS = sqrt(2(Pt-Ps)/Ps Gamma)

If you measure a higher Pt due to compressibility then yes you're going to need to correct it.
If you measure a CAS of 280kt at FL330, you're probably gonna want to correct something if you want to know your EAS.

In facts, you have to correct for compressibility if you use this above mentionned formula, simply because the formula is not valid for compressible flow.

For compressible flow, where a machmeter becomes useful, there are better suited formulas that you could use, and then you wouldn't need to correct for compressibility, because you avoided the need for it by choosing a better formula.

Then, I could agree that the answer marked as correct was actually wrong, i'm not in the head of the question's writers...
I was simply trying to find an explanation regarding this, without being able to tell whether I would answer 480 or 450 should this question appear on my official test (can we trust aviation exam or this other question bank when they tell us a surprising correct answer like this ?)

And from what you wrote after, I know you'll understand my message because you've written yourself the "better suited formula" I was talking about :)


Regarding the little technical details of real macheters, I really have no idea wheter early mach meters were able to mechanically reproduce this sort of complicated formula as written in wikipedia :
https://wikimedia.org/api/rest_v1/media/math/render/svg/e0b857bf039a0a00704bbe9b91782ff6024667fa
To begin with I don't understand how you could divide a number by another with a system of mechanical gears and hands, so I don't see how you would elevate this ratio to the power 2/7 :p

It is however very easy for modern ADC to use formulas as complicated as required thanks to IT.

[Alex Whittingham]

I think it is an old question, KayPam, now removed. I will research tomorrow.

[paco]

It would appear not to be in the database, Alex.

[selfin]

Alex,

1. On aircraft with Air Data Computers the ASI displays CAS
2. The Air Data Computer calculates CAS using variations of Saint Venant's formulae which assume compressible flow.

Those are equivalent statements and are, so far as I'm aware, required by SAE AS 8019 (and its successor 8019A) which is referenced by ETSO-C2d (https://www.easa.europa.eu/system/files/dfu/CS-ETSO.pdf) and US TSO-C2d (http://rgl.faa.gov/Regulatory_and_Guidance_Library/rgTSO.nsf/MainFrame?OpenFrameSet). I don't have a copy of the SAE Aerospace Standard but I believe the USAF counterpart is MIL-PRF-27197E (http://everyspec.com/MIL-PRF/MIL-PRF-010000-29999/MIL-PRF-27197E_42354/); table II uses impact pressure.

Because if #2 were true, the ADC would output what I know as EAS, CAS corrected for compressibility, although it still seems to be called CAS.

The second statement does not mean the ADC outputs EAS. If EAS is displayed as the primary airspeed by some ADC (fitted in a civil aircraft) then please provide a reference. The Saint-Venant solutions for compressible flow use impact pressure only. Constant CAS is equivalent to constant impact pressure. The Euler-Bernoulli solution for incompressible flow uses dynamic pressure only. Constant EAS is equivalent to constant dynamic pressure. The term "compressibility correction" is contentious in my opinion because it may suggest that EAS is more accurate than CAS. EAS clearly makes flight mechanics equations more tractable but it should be used with caution in the transonic regime.

KayPam,

Nothing prevents you from calculating a mach number, from speed and temperature, even in an incompressible flow.

The equation you would use was derived for a compressible medium and the only value Mach number can take in an incompressible flow is zero. Full understanding of the theory underlying the speed of sound proved surprisingly challenging in the history of science, defying such brilliant minds as Newton and Euler (it will be 2033 before all of Euler's work has been translated to English), and credit ultimately went to Laplace (Sur la Vitesse du Son dans l'air et dans l'eau (http://gallica.bnf.fr/ark:/12148/bpt6k65687125/f244.image), Annales des Chimie et de Physique 3, 238–241 (1816)). An excellent summary of the struggle to resolve this problem is given in Laplace and the speed of sound (http://www3.nd.edu/~powers/ame.20231/finn1964.pdf) by Bernard Finn (http://americanhistory.si.edu/powering/bios/finn.htm) who was, like John D Anderson, a curator at the Smithsonian Institution. Anderson provides a straight-forward derivation in Fundamentals of Aerodynamics, section 8.3, which is very similar to this this two-page derivation (https://upload.wikimedia.org/wikipedia/commons/e/eb/Introduction_to_Physical_Chemistry_Lecture_5_Supplement.pdf) plucked liberally from a search for "derivation of the speed of sound."

Nothing would prevent you from building a machmeter which would would exactly compute the mach from the following formula

An incompressible flow would prevent it. It isn't clear how you would propose to "correct" a zero Mach number. Such a Mach meter belongs in Eric Laithwaite's Multiplcation of Bananas by Umbrellas (http://www.gyroscopes.org/papers/The%20multiplication%20of%20bananas%20by%20umbrellas.pdf).

If you measure a CAS of ...

... then you've used an airspeed indicator using Saint-Venant's solution for an isentropic flow. That is, dynamic pressure cannot be measured in flight, and any incompressible flow solutions using a measured isentropic total pressure will be in error (including the fabled Mach meter you've proposed). The proposal amounts to misinterpreting useful information (impact pressure) as an approximation (dynamic pressure), followed by a "correction" to recover the original information. Such a scheme may work at low speed and low altitude but it is inconsistent and unnecessarily restrictive. Indeed it would have been simpler to have used Saint-Venant's solution directly in the first place to obtain CAS and Mach number, which is what modern ASIs and Mach meters do.

It is however very easy for modern ADC to use formulas as complicated as required thanks to IT.

That has already been achieved in purely mechanical instruments. There are numerous US patents outlining basic operations such as subtraction, division, and exponents can be dealt with by taking logarithms. You raise a very good question on precisely how Saint-Venant's solution can be implemented in analog ASIs and Mach meters.

[KayPam]

Sorry for having overcomplicated things in the following messages...
The equation you would use was derived for a compressible medium and the only value Mach number can take in an incompressible flow is zero.
First of all, I don't know why you're saying air would be uncompressible ? It is very much compressible when you squeeze it into a syringe or any kind of air pump or even a balloon. It also suffers waves of compression when sound travels through it. But even if we admit that LSS would be infinite in an incompressible flow :

Mach number is speed/speed of sound (TAS/LSS)

You're saying LSS is infinite in uncompressible flow... And you're probably saying that at low speeds the air is uncompressible.

Well, problem is sound does take time to travel even in still air. Even in water, a fluid that's almost uncompressible, sound takes time to travel.
So basically we have air at low speed : it's almost uncompressible so we can use Bernoulli's theorem. It is a tiny bit compressible so we can use the sqrt(gamma R T) = LSS formula. If speed increases, the LSS formula becomes "more correct" (according to you, you still have to explain why it would less correct at low speeds), but Bernoulli becomes less correct. And that's where the correction factor is needed.


But in any case, nothing prevents the engineer from performing computations that do not make perfect physical sense. One wants to compute a mach number as TAS/sqrt(gamma R T) ? No one can forbid him to do so.. Maybe tell them they're wrong but nothing more.

This is probably the sort of reasons why EASA ends up asking surprising questions like the one we saw.
Such a scheme may work at low speed and low altitude but it is inconsistent and unnecessarily restrictive.
It is probably the the same sort of reason that could lead to an EASA question like the one we saw yesterday.
Because except by using wrong formulas, I don't see why this 0.94 thingy would be required.


Correct equations allow us to forget about this kind of correction factor.

I was simply explaining the wrong physical reasoning leading to require a 0.94 correction factor.

[selfin]

KayPam,

Even in water, a fluid that's almost uncompressible, sound takes time to travel.

That means the bulk modulus K is finite. Increasing K results in higher speeds. In other words for an acoustic wave to have a finite travel time the volume of the material must change. The more pressure that is required to effect that change, the higher the speed.

In an incompressible flow (and medium) K is infinite which means the propagation speed is infinite, in turn reducing the Mach number to zero for all speeds.

So basically we have air at low speed : it's almost uncompressible so we can use Bernoulli's theorem. It is a tiny bit compressible so we can use the sqrt(gamma R T) = LSS formula. If speed increases, the LSS formula becomes "more correct" (according to you, you still have to explain why it would less correct at low speeds), but Bernoulli becomes less correct.

Assuming an incompressible flow at low speeds is reasonable because compressibility effects are negligible while more accessible equations describe flow behaviour very well. From a rational choice perspective the fidelity reduction is rewarded by more tractable mathematics so that the simplifying assumption is rational. In the same vein it would be irrational to introduce a Mach number to most incompressible flow treatments.

But in any case, nothing prevents the engineer from performing computations that do not make perfect physical sense. One wants to compute a mach number as TAS/sqrt(gamma R T) ? No one can forbid him to do so.. Maybe tell them they're wrong but nothing more.

To ensure consistency such a Mach number would need to be qualified as an artificial one. What would be the practical application of such a number?

Because except by using wrong formulas, I don't see why this 0.94 thingy would be required.

The factor in the question was irrelevant but there is insufficient information for us to determine why it was given.

[KayPam]

Yes, we finally come to a very good agreement !
What would be the practical application of such a number?
To ask a very strange EASA ATPL question :)

[KayPam]

I pass 33, 81 and 21 this morning.

However, there was an inconsistency in flight planning.

It said that contingency fuel should be 5% of the trip fuel, in the question.
However, we were given the cruise consumption of the aircraft and I was able to calculate that 5% of the trip fuel corresponded to 3 minutes of cruise consumption.

I thought there was a minimum of 5' fuel consumption for contingency ?

In this case, should we believe the question and consider that contingency is 5%, or apply the rules and consider it's 5' ?

Thanks

[paco]

Under AMC1 CAT.OP.MPA.150(b) Fuel policy it's the higher of 5% or 5 minutes at holding speed, so 5% would be correct in your case.

[KayPam]

5 minutes was higher than the 5% in this case.
And the question included the lower value (5%)

So should we disregard what the question indicates ?

[RedBullGaveMeWings]

A jet aeroplane is flying long range cruise. How does specific range / fuel flow change?

A Decrease / decrease
B Decrease / increase [marked wrong]
C Increase / decrease [marked correct]
D Increase / increase

Max range cruise speed is 1.32 Vmd and Long range cruise speed is 1.37 Vmd and is 4% faster and provides 99% of range compared to max range cruise speed. Am I missing anything?:confused:

[keith williams]

This is a very old question and you, like thousands of other students over the years, are making the mistake of comparing lrc with mrc. The question does not ask you to do this.

As time passes the aircraft becomes lighter, so the drag decreases. This increases the specific range and decreases the fuel flow.

The greatest surprise to me is that after all these years some of the questions banks do not explain this.

[RedBullGaveMeWings]

Ok, it makes logical sense.

[KayPam]

Well yes but the wording is a bit confusing :p
Adding a simple how does it change *over time* would be much clearer.

[keith williams]

The fact that many students make the mistake of assuming that the question requires a comparison between lrc and mrc, may be interpreted as evidence that the question is defective. But it is also possible that the real problem is the way in which many students are addressing the questions. There is actually nothing in the question to suggest that the lrc.mrc comparison is intended.

Students often make the same error with questions along through lines of *increasing flap angle in straight and level flight will?* many student pick answers such as *increase in lift* or *the aircraft will climb*. But these options ignore the clue *in straight and level flight*.

Although the lrc question has probably been deleted from the cqb by now, I believe that it should be retained in commercial question banks in order to illustrate the danger of misinterpreting questions.

[komail320]

I have question that really I can't find answer for it :( which is :

Q: During a pre-flight inspection it is noticed that the nose oleo strut extension is 2" instead of the
normal 6", but there is no sign of oil leakage. This indicates:

A. an internal oil leak
B. gas charge pressure too low
C. gas charge pressure too high
D. tire pressure too low

I am not sure but I think the correct answer is "B" :( PLEASE HELP ME

[keith williams]

Well, Let’s look at the options and see what we can deduce from them.

The length of the oleo is determined by the amount of oil and gas within it and the weight of the aircraft pressing down upon it. Any loss of oil or gas will reduce the length. Any excess of oil or gas will increase the length.

It is also worth noting that if the low gas charge pressure has been caused by a small gas leak, this leak will not be visually detectable. But an oil leak would be evident from the oil deposits in the area of the leak.

An internal (option A) leak would allow the oil and gas to mix, but would not change their total volume, so this would not change the oleo length. So option A is incorrect.

The length of the oleo is determined by the amount of fluid and gas which is held within it. So the tyre pressure (option D) is incorrect.

If the gas charge pressure is too high (option C) the oleo length will be greater than normal. So option C is incorrect.

If the gas charge pressure is too low (option B)the oleo will be shorter than normal, so this is the correct option.

[komail320]

Thanks , it was so useful

umm another question

When an aircraft employs static vents either side of the fuselage to minimize pressure errors it is
called:?

A.Static equalization
B. Pitot correction
C. Static balancing
D. Dynamic balancing

actually this is the first time hearing about static equalization :(

[KayPam]

Could anyone explain the difference in the usefulness of geocentric and geodetic (or geographic) latitude ?
I have remembered the 11.6 minutes maximum difference due to earth being oblate, and the difference in the definition, but I really don't see the point of geodetic latitude.

[paco]

It's because the earth isn't round - if it was, we would just use geocentric. Instead we have to draw a tangent at the horizon, then a line 90 degs to that, and the angle where it meets the horizontal at the centre of the Earth is the geodetic. The difference for our purposes is negligible, and if the earth were the size of a billiard ball you wouldn't notice any difference between it and the others on the table.

I don't see much point to it (for the exams at least), except maybe to say that the mathematical model inside the GPS (based on the World Geodetic Standard established in 1984) is different enough for you to be careful if your track takes you near to the boundaries of airspace or danger areas.

[selfin]

To amplify Paco's response, the orthogonality of geodetic coordinate axes simplifies the Euclidean norm. Using an orthogonal system is equivalent to setting the angle(s), between principal axes, in the law of cosines to 90 degrees. That is, coefficient F in the metric tensor (https://en.wikipedia.org/wiki/Metric_tensor) is zero when geodetic latitude is used instead of geocentric latitude, see equation 34 et seq in Thomas, P.D. (1952). Conformal Projections in Geodesy and Cartography (https://docs.lib.noaa.gov/rescue/cgs_specpubs/QB275U35no2511952.pdf). Special Publication No 251. US Department of Commerce.

A further disadvantage of using skew coordinates (i.e. geocentric latitude) is that each point on the height axis has a different latitude, except at the equator and poles.

[KayPam]

To clarify :

I'm assuming the geodetic coordinate axes at point P on the surface of the earth are :
- The perpendicular to the plane of tangency at P (up or down)
- On the plane of tangency a north vector
- A third vector determined with right hand rule
Whereas the geocentric axes at the same point would be
- A "vertical" on a line from center of the earth to P (up or down)
- A North vector
- An East or West vector

At this point I'm wondering why this second system of axis has a complicated Euclidean norm.
Maybe the North vector remained the same as in the first system of axes, then it means that the system of geocentric axes is not orthogonal (there is a small residual angle)

Is it correct up to this point ?

If yes, the other option would be to modify the North vector, to make it perpendicular to the line from the center of the earth. Then the North vector would go either into or out of Earth, and this is obviously a problem because it means that if you're travelling north or south at constant altitude then you actually have a vertical speed in this system of axes...

Regarding your last line : Up to reading your message I thought that GPS coordinates were actually the coordinates of the point, on the surface, of which we're vertical. But it's quite evident that it wouldn't work that way.

[selfin]

I'm assuming the geodetic coordinate axes at point P on the surface of the earth are ...

The curvilinear geodetic coordinates are simply longitude, geodetic latitude, and height. Your description is for an ENU or NED (https://en.wikipedia.org/wiki/North_east_down) frame.

Whereas the geocentric axes at the same point would be
- A "vertical" on a line from center of the earth to P (up or down) ...
Yes but this radius vector is not normal to the surface at P. Any other point Q along the surface normal to P has a radius vector with a different direction (i.e. different geocentric latitude). Contrast this with the fact that the geodetic latitudes of P and Q are the same. In other words, varying height has the major advantage of leaving geodetic latitude unvaried, which is a property of orthogonal coordinates.

At this point I'm wondering why this second system of axis has a complicated Euclidean norm.

Additional terms are needed to account for the change in geocentric latitude with changing height.

... I thought that GPS coordinates were actually the coordinates of the point, on the surface, of which we're vertical.

That is correct. WGS 84 coordinates are geodetic and therefore orthogonal.

[KayPam]

Selfin, you seem very knowledgeable :)

Another question or two about gen nav :
Why would twilight be shorter at the equator ? It seems logical to me that in winter, twilight should be longer and shorter in summer : earth's inclination about the ecliptic gives a "pseudo latitude" that's higher in winter and less in summer, providing higher insolation.
It's just as the ecliptic was a fictitious equator.
Does it work like that ?

A question in the database asked the date of the perihelion.
Why would this be of any importance when talking about gen nav ? The same course that states that perihelion/aphelion are of no importance to seasons and that their two values are very close (earth's orbit has very little excentricity)

Thanks

[aiyunke]

Does anybody know where I can find graphics that fit best with the "new" graphic questions in the Polish ATPL performance exam? I use aviation exam and mainly the SEP graphics match with the real test questions but the rest not. Does anybody know a better matching question bank? Thanks a lot!:)

[RedBullGaveMeWings]

What's so different or special with these graphic questions? I would like to know because I am sitting my Performance exam in a few days.

[selfin]

Why would twilight be shorter at the equator ? ... It seems logical to me that in winter, twilight should be longer and shorter in summer

There's an answer to that here (https://astronomy.stackexchange.com/questions/2408/why-is-twilight-longer-in-summer-than-winter-and-shortest-at-the-equinox), and in the final two pages of chapter 8 in Henning Umland's Short Guide to Celestial Navigation (http://www.titulosnauticos.net/astro/). Umland has neglected refraction (34 arc minutes) and one semi-diameter (16 arc minutes) of the sun, so a better value in lieu of his 24 minutes is 20.7 minutes, i.e. the centre sun travels through a local vertical arc of 5.17 degrees instead of 6 degrees during civil twilight.

A similar derivation is in WM Smart's Textbook on Spherical Astronomy (art 33, p 51). Corrections for altitude are available in the UK Air Almanac (http://astro.ukho.gov.uk/nao/publicat/ukaa.html), p 68 of 2017 edition; see also fig. 3 on p 69 for twilight duration in the Arctic depending on latitude and time of year. Despite the restricted latitudes in those figures the duration of twilight, on the condition of a sunrise existing (viz., upper limb touches the horizon), is shortest near the equinoxes.

Why would [the perihelion] be of any importance when talking about gen nav?

It's of absolutely no importance. Probably an inside joke among Vulcan navigators (à Le Verrier (http://adsabs.harvard.edu/full/1953ASPL....6..291E)).

you seem very knowledgeable

I'm afraid that is merely one of those internet mirages.

[KayPam]

Can anybody explain this ?
https://i.gyazo.com/735e4490a9888a169dd1dcddc5296a5f.png

I'm doubting this is the correct answer.. I don't see why it would be desirable to spiral towards the NDB

[keith williams]

The following two questions which are/were in the CQB require students to distinguish between HOMING and TRACKING. I have no idea where they got these definitions from (probably a 1950's RAF navigation course).

Question 1.
Which statement is correct for tracking towards an NDB in an area with constant wind and constant Magnetic Variation?

Correct answer given by the database:

The Relative Bearing of the NDB should be equal (in magnitude and sign) to the experienced Drift Angle

Question 2.
Which statement is correct for homing towards an NDB in an area with constant wind and constant Magnetic Variation?

Correct answer given by the database:

The Relative Bearing of the NDB should be kept 000 deg;

These two questions have caused much head scratching among students in the past and will probably continue to do so in the future.

[paco]

Homing is adjusting for wind by keeping the needle on the 0. This results in a phenomenon called the curve of pursuit (otherwise called bird-dogging) which takes you away downwind from an airway centreline - it was a common problem for bomb aimers, and why tracking is best, where you offset for wind and fly a straighter course.

[keith williams]

it was a common problem for bomb aimers

So the authors of the question didn't get it from a 1950's RAF course, but from a 1940's course!

[paco]

Yep! I saw the term in a Lancaster manual! But it is used in the US and Canada.

[KayPam]

Oh, I thought that questions about INS were old, I stand corrected :p

[paco]

I have a 1952 book that uses 1.23 for the VHF line of sight formula - yet the exams use 1.25.

[selfin]

Paco,

Thanks for the heads up on the path being a pursuit curve. The derivation from first principles in terms of groundspeed coordinates was not going well!

The solution in airspeed coordinates appears to be given on the Wikipedia page for a radiodrome here (https://en.wikipedia.org/wiki/Radiodrome), pushing the date back to at least the 1730s.

[keith williams]

I have just read the reference and two things occur to me.

1. The reference describes an object which is chasing another object which is itself moving. This is not the case with the navigation beacons.

2. If the hare is running directly away from the dog (which sounds like what any sensible hare would do) or directly towards the dog (a mad march hare possibly), the dog's path will be a straight line. This is of course the equivalent of the beacon being directly upwind or downwind of the aircraft.

[selfin]

Keith,

In the airspeed coordinate system the beacon has a velocity vector opposite to the wind vector.

Is it not correct that when the beacon is finally reached the aircraft velocity vector is parallel to the wind vector, in the opposite direction?

[keith williams]

My comments were not intended to be taken seriously

But

If we take the river-crossing scenario, if the food is some distance downstream and the current is sufficiently slow, the dog might actually be moving at right angles to the flow, or even in a somewhat downstream direction when it reaches the food.

Or in the case of the Hare running directly away from the dog, the entire path of the dog will be a straight line. Given the range of possible scenarios, we cannot say that the path will always be a spiral or even a curve.

I have not bothered to go through the maths in the link, but if it is correct, then presumably it will cater for the cases examples which I have described above.

[KayPam]

Paco,

Thanks for the heads up on the path being a pursuit curve. The derivation from first principles in terms of groundspeed coordinates was not going well!

The solution in airspeed coordinates appears to be given on the Wikipedia page for a radiodrome here (https://en.wikipedia.org/wiki/Radiodrome), pushing the date back to at least the 1730s.

Ahah, it's look like i've found someone more mathematical than me !
You're seriously interested in coming up with an equation to describe this kind of curve ?
I'm sure you would be familiar with the brachistochrone curve as well ?
You sound like some of my colleagues in higher education who went on to become math researchers.

[selfin]

For a pursuit starting with a non-zero crosswind component there will be at most only one instant when the headwind component is zero. The crosswind component will only be zero if the beacon is reached.

Nahin, referenced in the Wiki page on radiodromes, for a terrestrial coordinate system with the beacon at the origin and the wind acting along the positive y-axis, provides this function [eqn 1.2.8] for the radiodrome which he refers to as the "wind blown plane's path":

http://mathurl.com/mt4lwns.png

The wind speed is w and the true airspeed is v. At x = a, where the path crosses the x-axis, the heading is orthogonal to the wind. If n < 1, i.e. if the beacon can be reached, then the y-axis is an asymptote for y'(x). Therefore the beacon is reached directly from the downwind position unless it is initially approached directly from the upwind position.

KayPam,

That is a famous variational problem of similar vintage and one that took Newton half a day to solve.

[paco]

He obviously had trouble with the ADF as well :)

[keith williams]

Therefore the beacon is reached directly from the downwind position unless it is initially approached directly from the upwind position.

If I am understanding this comment correctly it means that:

Experiment 1. If the starting point is with the aircraft 10 km directly upwind of the beacon it’s final approach to the beacon will be from directly upwind.

Experiment 2. If we repeat the experiment, but with the aircraft displaced even a small distance (1 m, 1 cm or even 1 mm) to the left or right, it’s final approach to the beacon will be from directly downwind.

Is this really the case?

The limits of the resolution of the ADF system should not be a factor in this, because the same equations have been applied to other situations such as the dog and hare.

[eckhard]

Yes, I think that would be the case.
It's all about drift angle and not allowing for it.

[selfin]

Strictly in terms of the properties of the curve that is correct. The required rate of turn near the beacon will however often exceed what can be achieved. Arthur Bernhart discusses these curves in a set of papers here (https://www.researchgate.net/profile/Arthur_Bernhart).

[ghentpilot]

Hi there,
I'd need some feedback regarding the following question - I believe that the answer provided is incorrect.

QDB says that the correct answer is 277.5 m but I come up with 279.7 m applying the following formula:

Domain 1/2 width = 60m + 1/2 wing span + 0.125 D
Domain 1/2 width = 60 + 32.2m + (0.125 x 1500m)
Domain 1/2 width = 279.7m

Thanks

For aircrafts with wingspan over 60m the formula is: 90m + 0.125D; so 90 + 0.125*1500 = 277.5m

[paco]

The CAA response to post 1062:

Compressibility is typically considered at speeds above 200 kt CAS or 300 kt TAS.

However, it doesn't appear to be in the database anyway.

[Alex Whittingham]

Bless them, this is because the FAA documentation says compressibility need not be considered below 200KT (one assumes CAS but they don't specify) and 10,000ft (missed that bit out) and the CRP-5 and its RAF predecessors did not consider compressibility below 300KT TAS. The response is an awkward attempt to reconcile the two given speeds, but sadly misses the point that compressibility is not applied in the conversion between Mach number and TAS. Really.

[KayPam]

Hey,

Having just completed my 14th test a few days ago, I'd like to provide you with some feedback.
I started with a French School in mid december, so it took me a grand total of 6 months and one week. That's including the four total weeks I took to revise before sessions 1, 2 and 4.

My country does not seem to be willing to disclose your marks, but I had an average of about 95-96% with my school's final tests (they reused some questions from aviationexam and created some questions, to simulate new questions on the test day)

I spent half my time on the books and half my time on aviation exam.
This was done in parallel with my job, about 1 to 3 hours per day of work, every day during these six months.

About the subjects :
I started with :
Communications, air law, ops proc : just by heart learning, not so many new questions but they could be surprising since it's not easy to not miss anything.
Then, instruments, meteorology, perf and mass&balance : No idea if questions on test day were new or not because they could all be answered with reasoning. I like subjects like these.
Then AGK, PoF, flight planning. The latter two are like the four previous one : I liked them. However I feel AGK is the subject in which the level I reached is the weakest of all subjects : too many many systems and details to learn. Different type of fire detectors, seriously ?
To finish with, gen nav, radionav, human factors.
HF : easiest subject of all, took me 10 minutes to complete the test.
Gen nav : so many calculations, it makes me tired. Thankfully it's quite of an easy subject for an ex math-student like me.
Radionav : worst subject, by far. Many different themes and many new questions about ridiculous little details that no one cares about.. The subject in which I had the worst scores, immediately after having read the books. In the end I think I have a better command of this subject than AGK but it was a pain in the ass. Probably mostly because it was the last subject.

Now i'm on for some bureaucracy to get my official diploma.

I'm very glad it's over, I have no idea what I'm going to do with all the newly available free time.

[fly5N]

Hello, I came across a difficult PSR/PNR question while writing SACAA IR Exams and I need your help. Twice infact and all were similar.

I was given a PET Distance of I think 1200nm and distance from point A- B of 2300 NM if I remember correctly. Total fuel was about 15000kg and that was all. No speeds or endurance was given or wind. I was then asked to calculate PSR from point A to B.

[Piloto Maluco]

Hi there

I started studying from the syllabus learning objectives, means that I learn specifically what they are willing us to know, line by line.
Do you think that's a good technique to pass the ATPL theory exams?! Or I should only study from my notes/books/cbt,... ?!
I am quite worry as the official exams are fastly coming up...

Cheers

[paco]

In that case you will learn to navigate before you know about Lat & Long......

Not a good idea that. The LOs were never meant to be a syllabus as such.

[delaneyslad]

Hey Pacol, then why put LO's in? What's the point of covering LO's if they don't (at the very least) cover the syllabus? Can you point me to the syllabus per se? That way I won't have to worry about the LO's

[paco]

There are no syllabuses, that's the rub! They specify an LO, meaning what a student is expected to know after the studying then leave it up to the schools to guess to what level they should teach, thus creating their own syllabus. The only reference to levels or even sources of knowledge is in the question itself, which nobody should be able to see.

[Rusty1983]

Hello, I came across a difficult PSR/PNR question while writing SACAA IR Exams and I need your help. Twice infact and all were similar.

I was given a PET Distance of I think 1200nm and distance from point A- B of 2300 NM if I remember correctly. Total fuel was about 15000kg and that was all. No speeds or endurance was given or wind. I was then asked to calculate PSR from point A to B.

Hello maybe I can contribute to the question because I had a similar one,

I got during my IR rating exam an interesting PSR\PNR question where I am going right now crazy to get to a solution. I asked now already everyone at flight schools, looked in different forums, but could not find an answer. I even asked already the authority, but they were not authorized to help me...

Hopefully you guys can help me here with a solution.

The question simple:
- distance A to B 2000NM
- TAS, wind, endurance constant
- 5000kg fuel (+500kg extra)
- PET from A 1200NM

what is the PNR\PSR?

answeres: I cant remember the values any more but for the approach process it should be not an issue. There were 5 answeres and I chose c.) which was 1200NM (same as PET and which is wrong).

My approach was the following, due to the fact that there are to less variables for calculations, no given fuel flow,..., it's more a theoretical question. TAS, wind, endurance are constant, so I thought it might be one of the rare cases where PET=PSR, which leads me to another question:
- Can PSR be equal PET and\or is PSR always more\less then PET?

I assume that when there is no wind and PET is exactly half of the distance, is there also PSR half of the distance or more\less?

In the example above I know I have headwind doe to the fact that PET is closer to B, so if PSR is maybe always more then PET then there might was only one answer which was more then 1200nm...

Or maybe I am completely wrong...,
anyways, appreciate your help in this case!

[Rusty1983]

And maybe another question which is right now a little bit a pain in my ass, where even after discussion with mechanics I came to no solution.

Which TWO instruments are adversely affected by an electrical power failure in flight?:
A. ASI
B. Magn. compass
C. DI
D. AI
E. CDI

A and b definitely not. I would have chosen c,d,e but the question said TWO, and now i am strungling. The discussion with the mechanics ended up that it's c and d, but also a cdi could be affected depending on an old or new one...

So hopefully someone here knows the right answer and can explain why.

Appreciate it, thx

[paco]

The compass is swung with all the electrics on, so if they fail, the deviation card will be inaccurate.

[Rusty1983]

Thx for this true statement, unfortunately helps not for the answer to the question.

[WASALOADIE]

IIRC, An AI can be either air driven or electrically driven so therefore is not entirely susceptible to a power failure. Is that the possible reasoning ?

[Sonbra]

Probably a super stupid question but how do you know if it's starboard or port drift? When you are calculating headings on the CRP 5 (chapter 8 Oxford ATPL navigation). Example: the heading is 060 and the track is 065 which means you have 5 degrees starboard drift. I understand how they got the 5 and all that. But how do I know if it's port or starboard? Any rules of thumb?

[RichardH]

There is no such thing as a stupid question Sonbre! Unless it's "what colour is the green navigation light"?

Anyway assuming you are using wind-down method (one reason it's used is it shows drift in the correct sense). Your heading will be at the top and your pencil dot will be usually be on the left or right hand side of the CRP5 centre line.

In your example heading of 060 is at the top and the pencil dot will be on the 5 degree right drift line. If it's on the right then this is starboard drift, if your pencil dot is on the left then this is port drift. Remember that wind ALWAYS blows from heading to track.

These phrases like several aviation ones date back to the days of sail, (like some of the dire questions) in the EASA exams they are more likely to use the phrases left or right drift.

[keith williams]

Port means your left side and starboard means your right side. So in the example you have given, you are looking along your 060 heading, but actually moving in the direction 065. This means that you are drifting 5 degrees to your right (starboard) side.

[Rusty1983]

IIRC, An AI can be either air driven or electrically driven so therefore is not entirely susceptible to a power failure. Is that the possible reasoning ?

Which means the correct answeres would be the cdi and the di...but also not 100%sure.
I love it when the authorities give you not enough infos for solving a question.:ugh:

[Alex Whittingham]

You have to love the examiners' simplicity, not specifying the type of aircraft. Or maybe they did and its not included in the feedback?

Which TWO instruments are adversely affected by an electrical power failure in flight?:
A. ASI
B. Magn. compass
C. DI
D. AI
E. CDI

On an aircraft like the older 747s with servo driven main instruments the answer is "all" but one has to assume something so I will assume they are talking about a light aircraft. On such an aircraft:


the ASI has no power supply, it is not affected
the direct reading magnetic compass has no power supply, but its reading is affected because the deviation correction is done with all power supplied, and therefore it may be incorrect if the electrical power fails. If the compass is remote reading it will require power and therefore be affected.
the DI is air driven
the AI is air driven
the CDI (if it has one) is powered


and therefore I would go for B and E.

[Rusty1983]

Thx for the feedback. No, the question is like its mentioned. There is nowhere an a\c type or heavy\light a\c,...

And you are right, found also some interesting stuff in inet:
Link: http://www.cfinotebook.net/notebook/...scopic-systems

With this in mind, the correct answers are:
Magnetic compass and CDI

And I was always so sure that the magn. Compass is definitely affected but that much that it could be adversely affected (only a few degrees but still working), but I think this is the reason why the authorities use these words. Only to trick you around and make you doubting your thoughts, even when they are right.

[momo95]

Can someone be so kind as to explain briefly the working principle behind spring tabs?

Also, would I be correct in saying that we have

External/Control Balance Tabs

- basic elevator tabs (controlled by a wheel/flick switch),
- Balance/Anti Balance tabs (not moved by pilot ... move with control surface)
- servo tabs
- spring tabs

Internal Balance Tabs

- control horns
- insert hinges


The above being all that is necessary for the ATPL (POF) ... this shouldn't be a difficult topic but it's something I'm slower to get to grips with than others. I have used the BGS ATP extensively but unfortunately some things are still not clear for me. Any help is appreciated :)

[whiskey1]

Need a minimum of 4 satellites for a 3d fix from GPS.
Latitude, Longitude, Altitude and Time all need to be determined.

[vollkornbrot]

Hi,

can anyone help with this question please? :-)

An aircraft follows a true course of 140°(T) for 1 hour 42 minutes between 45°N 070°W and 42°N 062°W. Calculate how much drift (and in what sense) will be observed on an uncorrected DGI.

Solution -24°

[KayPam]

The drift is the heading error.
It is due to two things :
- earth rotation
- moving along a parallel of latitude
You've got here :
1.7h * 15°/h = 25.5°
8° of longitude variation
Multiply that by sin(meanlat) which is about 0.7
You'll find something very close to -24°

Just take a few minutes to imagine the situation and confirm that both signs are negative.

[Sonbra]

Can anyone help me with this?

The book says: Calculate the airspeed in knots for each highlighted coefficient of lift.
I don't need the answer for a specific number I just want to know how to do it. What formula to use and so on.
Thanks;-)

[keith williams]

Sonbra,

You have not provided many clues, but I suspect that the equation you require is the lift equation.

Lift equals Cl 1/2Rho Vsquared S.

If we keep Rho and S constant then to maintain constant lift for any changes in Cl we need an equal and opposite change in V Squared.

So for example if we half Cl from say 0.5 to 0.25, then we must double V squared. To achieve this we would multiply V by the square root of 2.

[Sonbra]

Thanks! I do understand the formula. But I don't get the right answer!:ugh:
The CL is 0.384, the Newton of lift is 588600N (60000X 9.81) the wing area is 105 and the density is 1.225 kg/m3

The answer supposed to be 300knots. But how?

[keith williams]

You must ensure that you use consistent units throughout the calculation. You have Newton, and metres, so you need to convert your 300 knots into metres per second. Hopefully your books have stated what conversion factor to use, but if not you should try using 0.514 m/S per knot, which is a reasonable approximation.

[Rusty1983]

Hello maybe I can contribute to the question because I had a similar one,

I got during my IR rating exam an interesting PSR\PNR question where I am going right now crazy to get to a solution. I asked now already everyone at flight schools, looked in different forums, but could not find an answer. I even asked already the authority, but they were not authorized to help me...

Hopefully you guys can help me here with a solution.

The question simple:
- distance A to B 2000NM
- TAS, wind, endurance constant
- 5000kg fuel (+500kg extra)
- PET from A 1200NM

what is the PNR\PSR?

answeres: I cant remember the values any more but for the approach process it should be not an issue. There were 5 answeres and I chose c.) which was 1200NM (same as PET and which is wrong).

My approach was the following, due to the fact that there are to less variables for calculations, no given fuel flow,..., it's more a theoretical question. TAS, wind, endurance are constant, so I thought it might be one of the rare cases where PET=PSR, which leads me to another question:
- Can PSR be equal PET and\or is PSR always more\less then PET?

I assume that when there is no wind and PET is exactly half of the distance, is there also PSR half of the distance or more\less?

In the example above I know I have headwind doe to the fact that PET is closer to B, so if PSR is maybe always more then PET then there might was only one answer which was more then 1200nm...

Or maybe I am completely wrong...,
anyways, appreciate your help in this case!


Hello,
Unfortunately I got no response on the question above, but maybe someone have a clue of a new one that I got at my last exam.

Even this one I was not able to figure out or find any solution hints in the internet. I really like to meet these Sadists who create such questions!

Dist. A - B = 1200Nm
PSR is 84% of AB
PET is 60% of AB
Endurance = 8h24mins

What is the Groundspeed from PSR to A?

I am going already crazy with these questions...everyone I ask, no one has a clue

[Alex Whittingham]

Let's have a go...

We know distance to PET = (DH)/(O+H), substituting..

(60/100)*1200 = (1200*H)/(O+H) therefore
(60/100) = H/(O+H) and, if we wanted to, we could find O in terms of H

also distance to PSR = EO'H/(O'+H), using O' for distance out because groundspeed out to PSR/PET may not be the same as groundspeed on from PET (engine fail case etc. - similar arguments can be presented for H not being the same as H')

substituting...

(84/100)*1200 = (8.4*O'*H)/(O'+H)

Now at this point, if O and O' were the same, we could substitute our values for O in terms of H from the first formula and find a numerical value for H. We don't know this is the case, though, and so here I stop. My comments would be (i) that this is not a fair test of knowledge of formula, rather it is a test of maths and (ii) the solution relies on an assumption that we cannot make. All comments and corrections welcome.

Could you tell us which exam this appeared in, please, and which examining State?

[Sonbra]

I´m kinda stuck here. :confused:
In a 45 degree turn L=1/0.707 = 1.41 Okay I get that, but how do they get 0.707?

I get that the the formula is: cos Ø= ADJ(1)/HYP(L)
transposing this formula gives, 1/ COS Ø

How do I get the 45 degrees to become 0.707?

Anyone that wanna help me and explain this step by step?
:ugh:

[Alex Whittingham]

What value are you getting for cos 45 degrees?

[paco]

For the question above, are we talking about PSR (in which case reserves are involved) or PNR (with none)?

Edited: it would appear that the terms PSR/PNR in EASA questions are interchangeable - in other words all endurance values given in questions are safe values and can be used directly.

As the PET always moves into wind, shirley you would have a tailwind going back to A? (if you draw a diagram, you will see that the PET is past the midway point, more towards B so, by definition, there has to be a headwind outbound).

[smthngdffrnt]

What happen with the left wing AoA when rolling to the right?

Sometimes I read that the AoA is higher because of the deflection of the aileron downwards and every now and then is lower attending to the rolling effect -lift vector tilts back due to the change of the relative wind-.

Can anyone help me to understand this possible contradiction?

[paco]

Both I suppose - for example, in a helicopter, a rising rotor blade has a reduced angle of attack because of the change in relative airflow, a similar situation to a rising wing, but helicopter blades don't have ailerons, so the AoA will also increase when they are used.

[Rusty1983]

Let's have a go...

We know distance to PET = (DH)/(O+H), substituting..

(60/100)*1200 = (1200*H)/(O+H) therefore
(60/100) = H/(O+H) and, if we wanted to, we could find O in terms of H

also distance to PSR = EO'H/(O'+H), using O' for distance out because groundspeed out to PSR/PET may not be the same as groundspeed on from PET (engine fail case etc. - similar arguments can be presented for H not being the same as H')

substituting...

(84/100)*1200 = (8.4*O'*H)/(O'+H)

Now at this point, if O and O' were the same, we could substitute our values for O in terms of H from the first formula and find a numerical value for H. We don't know this is the case, though, and so here I stop. My comments would be (i) that this is not a fair test of knowledge of formula, rather it is a test of maths and (ii) the solution relies on an assumption that we cannot make. All comments and corrections welcome.

Could you tell us which exam this appeared in, please, and which examining State?


Hi and thanks for the answer.
I got it at the new IR Exam in South Africa. (This exam is way out of anything from the difficult level I wrote in all my CPL exams). I thought I understand FP got at my PPL and CPL exam 100%...but this at the IR Exam is way another level...I wrote it now 3 times and always got an PET/PSR question which was way out of that what you learn in the theory.
Problem is it is marked with 3 to 4 points in the exam, which is quiet heavy.

Since I am writing this exams (April 2017, no one from the chopper guys in whole South Africa passed this exam and the pass rate at the fixed wings is what I saw 0-2%).

Since the CAA changed the syllabus and the questions since mid 2016 the DFE's and instructors running massively against this new exam, because only 4 of 96 DFE's passed the exam itself. But the authority say, it's ok...

[Alex Whittingham]

Terrible, Have just done a rough calculation assuming all Os and Hs are the same and get 300KT. Was that an answer?

[paco]

after consultation, simultaneous equations...

(D x H)/(O + H) = PET for a tail wind home, with O=2/3xH.

Plugging that into PSR=(ExOxH)/(O+H) and rearranging comes up with H=300

You're right - a completely impractical question

[Rusty1983]

Terrible, Have just done a rough calculation assuming all Os and Hs are the same and get 300KT. Was that an answer?

i came during the exam due to my good math skills to an answer with weird calculation of unknown factors ended up at 200kts, which was an answer...unfortunately the wrong one.
There were 5 answers, I dont know if 300 was one of them, but Ill check next time, when and if Ill get it again.

But one thing is not logical for me in this calculation. I know that assuming O and H the same the calc would be possible, but what makes a knot in my head is, that O and H cannot be the same, due to PET is after the midpoint, which means I have a headwind in going out and a tailwind in going back. The only thing what I assume now is, that I use now O or H and make out of the two an equation with only one variable and put this then in the big PSR formula....but then I get a few knots in my head...

[Alex Whittingham]

I haven't explained myself properly. In the formula for PSR 'O' stands for groundspeed out to PSR. In the formula for PET 'O' stands for groundspeed onwards from PET. Assuming TAS and wind components are the same of course then these two speeds would be the same. But this is not necessarily the case, for instance we could be calculating an engine failure PET where the speed out to PET is based on an all engines TAS and the speed either onwards or back from PET is based on an engine out TAS, perhaps at a completely different height. Its a common mistake to assume that, because both formulae use the same letter 'O' that the two formulae can be related. The examiner has made this mistake.

I did not mean that O=H, I meant I assumed, as the examiner seems to have done, that the O in the PSR formula is exactly the same as the O in the PET formula. Likewise in order to get the answer I have assumed that H has the same value in both formulae, and it need not have.

[paco]

I think the only way to do that question suitable for a pilot is to take the total PSR distance flown (2016), divide it by the endurance (8.4) to get a mean speed of 240 kts - closer to 200 than 300 because more time is spent at 200. Then look for an answer that is faster than that. The PET I think is a red herring.

[Runway37]

Whilst watching the demo for the ATPL exam format (Quadrant), I have come across this question, which seems very ambiguous to me.

Does anybody have any idea on what they would write for this type of question?

Screenshot attached.

[paco]

My understanding is that there are no essay questions for the UK exams

[Kakaru]

Hi folks,

Met question from one of the Q banks.

Carburettor icing may be expected to occur in clear air in the following conditions:

a. relative humidity 20%, OAT +30 C and climb power set.
b. relative humidity 40%, OAT +20 C and descent power set.
c. relative humidity 30%, OAT +30 C and descent power set.
d. relative humidity 20%, OAT +30 C and cruise power set.

C is the correct answer, but why not B?

PS. I can't attach the well known carburettor icing graph here since I have less then 10 posts.

Thank you all in advance!

[Alex Whittingham]

Beats me, you should ask the question bank provider. I would have chosen B

https://www.atsb.gov.au/media/5747607/rId34%20Carburettor%20icing%20probability%20chart_500x353.jp g

[Kakaru]

Thanks Alex. Looks like this needs to be appealed / reported. My logic was that the higher relative humidity would be a deciding factor here (even if you don't know the graph) and according to graph, B is closer to the "red" zone.

[cavok_flyer]

A SOB (Senile Old :mad:) from the LBA once told me that the answer will always be the one with the higher temp, since they wish to make the point that carb-icing can happen even at higher temperatures.

[rtraveler]

Hello guys! I am preparing for Atpl questions, i am already studying from different resources, i wonder if anybody of study from atplquestions.com website?

[ersin737]

I have been study for Performans and one question make me very nervious because there is two answer I have to make a sure which one is the correct. this question coming from new EASA data base ECQB4.

(For this question use annex ECQB-032-023-v2015-04 or CAP 698 Figure 2.3). Given the following information, what is the Rate of Climb? Pressure Altitude: 10000 ft ISA conditions Aircraft Mass: 3400 lb Full Throttle, 2700 RPM”

A-) 640 ft/min
B-) 500 ft/min
C-) 720 ft/min
D-) 1180 ft/min

I founded 640 ft/min however the question bank show me the corect answer is 720 ft/min. does anyone explain this question.

[ersin737]

Most of the question from new ECQB4 database, I know who made it this web site, you can use. also aviationexam my favorite web site.

[selfin]

That question is so poorly worded that it ceases to be meaningful. It seems to ask for the final forward azimuth of a geodesic (great circle) between the given stand- and forepoints. On the WGS84 ellipsoid this angle is 249 deg (east of true north). See `fazi2` in Charles Karney's GeodSolve utility here (https://geographiclib.sourceforge.io/cgi-bin/GeodSolve?type=I&input=53d50%27N+006d55%27E+53dN+003dE&format=g&azi2=f&unroll=r&prec=0&radius=6378137&flattening=1%2F298.257223563&option=Submit) (add 360 degrees to negative values). Chris Veness's implementation of the older Vincenty algorithm for the inverse solution can also be used, here (https://www.movable-type.co.uk/scripts/latlong-vincenty.html). In practice you'd lay a line down on a Lambert conformal conic projection, e.g. UK CAA VFR Air Chart, and measure the local direction at the forepoint 53N 003E.

"Final true track" is a mongrel construct and should be avoided. "Track" is not explicitly defined as a loxodrome (rhumb line), or a geodesic (great circle), in any major text on navigation. Paths with constant true track values are always loxodromes so the qualifier "final" is redundant.

[Alex Whittingham]

Feedback suggests there's a bit missing from the Q above, rhumb line track is given as 250 deg.

Convergency = change of long x sin mean lat
= 4 deg x sin 52.5 deg (roughly)
= 3.2 deg (roughly)
conversion angle is half this, 1.6 deg
subtract from 250 to get 248.4 deg (roughly)

Otherwise what selfin said ^^^^, but by the standards of EASA exams its an OK question.

[SkillsToBurn]

how do i calculate the cross wind of - Wind 100/30kts with runway of 340?

[ab154]

Sin(difference between runway orientation and wind direction) x windspeed.

In this case Sin(120)x30 = 26 knots

[ab154]

Try going around the other (shortest) way.

340 + 20 gives you 360/000 then you have to add 100 to get to 100 degrees.

For what its worth, you can use Sin(240)x30 and the answer is -26 knots. If you remove the minus sign you end up with the same answer anyway.

[dan1165]

340 ° - 180° = 160 °
Wind from 100 ° -> sin 60 = 0,8
0,8 x 30 = 24 kt

[Sepp]

Am I missing something, difference between Runway and wind is 240 degrees?

340 - 100 = 240?

Where did 120 come from?

Negan, the wind is from the direction of 100deg, not the direction that is 100 deg away from 340.

The angluar distance between 340 deg and 100 deg is 120 degrees.

If you still can't see it, look at the compass rose on the wind side your CRP-5 (or equivalent) and count the 10-degree divisions between the two directions.

[SkillsToBurn]

I was wondering if anyone could help me with these 3 questions from comms exam that came up this week?

IFR Comms

Q. Isolated CBs have been forecasted along a route. An aircraft flying in a layer of nimbostratus sees a lightning flash. What actions should the pilot take?

a) Declare a PAN as he has flown into a storm.
b) Transmit an AIREP because the CB is embedded. <<<<<<<
c) Report the Lightning at the next waypoint, which is only 15 mins away.
d) Nothing, because the CBs have been forecast.

Q. For which of the following transmissions should each number be spoken individually?

a) Climb to 2500 feet
b) Visbility 4000 metres
c) QNH 1000 <<<<<<<<<<<<
d) Overcast 1300 feet

Q. Aircraft may be required by ATC to make airborne reports of air temperature, wind and turbulence. Aircraft not equipped with data link equipment are:

a) Exempt to make regular reports, unless specifically requested by ATC <<<<<<<<<<<<
b) Required only to report once, immediately after being requested
c) Are required to make regular reports, irregardless of position.
d) Are only required to make reports upon passing reporting points, not more than 30 mins every time.

[Alex Whittingham]

A note on the first question, NS is associated with warm fronts and stable air and its very rare to get embedded CB in NS, although it can unusually happen when the warm sector is very unstable (unusual) and initial frontal lifting sets them off. One would have hoped the forecasters would have picked this up and included the possibility of CBs embedded in the NS, but apparently not and only ISOL CB were forecast not ISOL EMBD CB, and therefore the answer selected is correct.

[SkillsToBurn]

Thank you Alex!

[Alex Whittingham]

At least I hope!

[SkillsToBurn]

I was torn between B and D in all honesty! Since the CB has already been forecast, if it was in a practical flying sense then you would have aircraft continuously calling up along the route/Airway, reporting the CB/Lightning!!

[paco]

In case anyone is interested, in answer 2, this is straight from CAP 413:

2.13 When transmitting messages containing aircraft callsigns, altimeter
settings, flight levels (with the exception of FL100, 200, 300 etc.
which are expressed as ‘Flight Level (number) HUN DRED’),
headings, wind speeds/directions, pressure settings, airspeed,
transponder codes and frequencies, each digit shall be transmitted
separately;

[Alex Whittingham]

...not that CAP 413 is the reference document for this subject. It should be the much more vague ICAO Annex. However I agree with you.

[paco]

Bryan Johnson told me way back when that the ICAO document is the only one that follows a national document, in this case CAP 413...... Put another way, 413 is edited first. Go figure.

[mattria]

Hello,

For the ATPL theory, are question banks (like Bristol Ground School) a more valuable tool than Oxford's CBT? I was looking over some Human Performance questions and the banks don't get into as much detail as the CBT. For example, the questions touch the surfaces of Dalton's Law, yet the CBT seems to be too in-depth.


Thanks,

[danne9595]

Think the best solution is to use them both!
Personally i use Aviation exam instead of bristol but I also use the CBT to get some deeper understanding, Question banks is ideal to know what you are supposed to focus on! gl! =)

[MaverickPrime]

Why can the following METAR not be abbreviated to CAVOK?

DLLO 121550Z 31018G30KT 9999 FEW060TCU BKN070 14/08 Q1016 TEMPO 4000 TS=

(Aerodrome elevation 1000 ft, MSA for sector 000-190° 5800 ft, for sector 190-360° 7300 ft.)

Accord to QB correct answer is:

Because the cloud base is below the highest minimum sector altitude.

I'm a bit lost with this one. As far as I understand you don't add the elevation to MSA as MSA is 'altitude' above sea level? If you do add the elevation then I understand how they get this answer.

Practically speaking, if the MSA is 5800ft MSL then I don't think I'd fly 200ft below TCU, so it is common sense at that.

[hvogt]

The term 'cloud base' in the answer given as the correct one is slightly misleading as the definition of CAVOK does not refer to any cloud base. However, the conditions for CAVOK are not met since (1.) there are clouds below 1 500 metres (5 000 ft) or below the highest minimum sector altitude, whichever is greater, and (2.) there is towering cumulus.

[MaverickPrime]

The term 'cloud base' in the answer given as the correct one is slightly misleading as the definition of CAVOK does not refer to any cloud base. However, the conditions for CAVOK are not met since (1.) there are clouds below 1 500 metres (5 000 ft) or below the highest minimum sector altitude, whichever is greater, and (2.) there is towering cumulus.

TCU was one of the options, but it was marked as incorrect, which threw me further.

I guess the examiner was thinking along the same lines as Negan.

I got this from the BGS bank, but I didn't take note of the question ID.

[Alex Whittingham]

ICAO Annex 3 says:

2.2 Use of CAVOK
When the following conditions occur simultaneously at the time of observation:
a) visibility, 10 km or more;
Note.— In local routine and special reports, visibility refers to the value(s) to be reported in accordance with 4.2.4.2 and 4.2.4.3; in METAR and SPECI, visibility refers to the value(s) to be reported in accordance with 4.2.4.4.
b) no cloud below 1 500 m (5 000 ft) or below the highest minimum sector altitude, whichever is greater, and no cumulonimbus;
c) no weather of significance to aviation as given in 4.4.2.3 and 4.4.2.4;

...and our interpretation is that TCu is not CB

[MaverickPrime]

ICAO Annex 3 says:

2.2 Use of CAVOK
When the following conditions occur simultaneously at the time of observation:
a) visibility, 10 km or more;
Note.— In local routine and special reports, visibility refers to the value(s) to be reported in accordance with 4.2.4.2 and 4.2.4.3; in METAR and SPECI, visibility refers to the value(s) to be reported in accordance with 4.2.4.4.
b) no cloud below 1 500 m (5 000 ft) or below the highest minimum sector altitude, whichever is greater, and no cumulonimbus;
c) no weather of significance to aviation as given in 4.4.2.3 and 4.4.2.4;

...and our interpretation is that TCu is not CB

Thanks Alex, that clears it up for me!

[Ronaldsway Radar]

And purely for reference, but here is the extract from EASA consolidated SERA, effective October 2017:

SERA.9010 Automatic Terminal Information Service (ATIS)
(b) ATIS for arriving and departing aircraft
...
(13) visibility and, when applicable, RVR (*) and, if visibility/RVR sensors related specifically to the sections of runway(s) in use are available and the information is required by operators, the indication of the runway and the section of the runway to which the information refers;
(14) cloud below 1 500 m (5 000 ft) or below the highest minimum sector altitude, whichever is greater; cumulonimbus; if the sky is obscured, vertical visibility when available; (*)
...
(*) These elements are replaced by the term ‘CAVOK’ when the following conditions occur simultaneously at the time of observation: a) visibility, 10 km or more, and the lowest visibility not reported; b) no cloud of operational significance; and c) no weather of significance to aviation.

And from the same document:

‘cloud of operational significance’ means a cloud with the height of cloud base below 1 500 m (5 000 ft) or below the highest minimum sector altitude, whichever is greater, or a cumulonimbus cloud or a towering cumulus cloud at any height.

The term 'weather of significance to aviation' I could not find a definition for within SERA, and wonder how this differs from the previous term. I believe that ICAO Annex 3 defines both terms.

Clear as mud :)

[Alex Whittingham]

Fascinating, thank you RR. I will ask the CAA what document we are meant to be teaching to!

[byronmc]

Hey everyone

I'm a little stuck with this :ugh:Any help would be appreciated

Need 2.6% climb gradient, currently have 2.8% at 110000KG. What will the new
MTOW to have 2.6%?

[superflanker]

Hey everyone

I'm a little stuck with this :ugh:Any help would be appreciated

Need 2.6% climb gradient, currently have 2.8% at 110000KG. What will the new
MTOW to have 2.6%?

It's just a proportions problem. As weight increases, climb gradient decreases (this part is obvious). So the answer must be higher than 110000Kg.

How much is that increase? 2.8/2.6 = 1.077 --> 110000 * 1.077 = 118470Kg aprox.

[byronmc]

It's just a proportions problem. As weight increases, climb gradient decreases (this part is obvious). So the answer must me higher than 110000Kg.

How much is that increase? 2.8/2.6 = 1.077 --> 110000 * 1.077 = 118470Kg aprox.

Thank you very much Superflanker:ok:

[SkillsToBurn]

Alright Folks, just been scanning the recent feedback and have come across this question which has been cropping up a few times.

You are in a power on decent with a glide angle 30 (degrees)

Weight - 11000
Lift- 10500
Drag - 8500

What is the "thrust vector" in KG? (Type in)

Obviously the variables will change in regards to the Angle, weight etc, so its best I know the method of calculation.

If anyone could shed more light on this, would be much appreciated.

[Alex Whittingham]

Resolve the forces along the longitudinal axis of the aircraft. In one direction we have thrust plus the 'thrust component of weight', W sine theta where theta is the descent angle. In the other we have drag so:
T + W sine theta = D
or T = D - W sine theta.
substituting,
T = 8500 - (11000*0.5) = 3000 Kg. How did I do?

[SkillsToBurn]

Thanks again Alex!

Unfortunately I have no answer spread, so ill have to take your word for it ;). There have been many different ways the examiner is asking this question, with different numbers etc feedback from other candidates show some very varied answers! So I thought I would ask someone who actually knows what they are doing. They are also asking a similar question but in the climb!

NB :- Although feedback is good, should be taken with a pinch of salt!

[atpl95]

hey folks
Can someone help me understand how to solve problems like this? Im kinda stuck:ugh:

An NDB is located in position 4500S 14500W. Your ADF indicates relative bearing 292° to this NDB. The convergence between the NDB and your DR position is 10°. Variation is 11°E at the NDB and 15°E at your DR position. Your magnetic heading is 295°. Using an Equatorial Mercator chart, you shall plot the position line, correcting for the convergence.What is the true direction of the plotted line of position, as plotted from the NDB?
What will, based on the plot calculations, the initial magnetic heading be if you decide to fly the great circle track from your present position to the NDB?
You calculate to get 7° right drift along the track to the NDB

[Alex Whittingham]

QDM is relative bearing plus magnetic heading, 292 + 295 = 587, minus 360 to make sense of it = 227 deg mag

true brg to is QDM +/- variation at the aircraft = 227 + 15 =242 deg

This represents the initial true great circle track aircraft to beacon, as radio signals follow the great circle path.

On Mercator charts, however, one has to plot the equivalent rhumb line because great circle paths are shown as curves on this particular archaic form of chart, it is optimised for marine navigation following lines of latitude east/west which show as straight rhumb lines (see C Columbus and 15th century navigation).

The angular difference between the great circle track to the beacon and the rhumb line track to it is half the convergency, 5 degrees. This is called the conversion angle. The outstanding question is whether this should be added to or subtracted from the great circle path. This is decided by remembering that great circles always lie to the poleward side of the equivalent rhumb line and therefore, for a generally westerly great circle bearing of 242 in the southern hemisphere the equivalent rhumb line would be 247 degrees.

If you wanted to plot the rhumb line from the beacon to the aircraft as the question asks you would plot the opposite, 067 degrees.

The second part of the question asks the initial magnetic great circle track to fly. We have already established the QDM is 227, 7 degrees of right drift would require a mag heading of 220 degrees.

I'm guessing this is a school question. How did I do?

[dook]

I'm guessing this is a school question.

Just like centuries old spoof rubbish.

[atpl95]

QDM is relative bearing plus magnetic heading, 292 + 295 = 587, minus 360 to make sense of it = 227 deg mag

true brg to is QDM +/- variation at the aircraft = 227 + 15 =242 deg

This represents the initial true great circle track aircraft to beacon, as radio signals follow the great circle path.

On Mercator charts, however, one has to plot the equivalent rhumb line because great circle paths are shown as curves on this particular archaic form of chart, it is optimised for marine navigation following lines of latitude east/west which show as straight rhumb lines (see C Columbus and 15th century navigation).

The angular difference between the great circle track to the beacon and the rhumb line track to it is half the convergency, 5 degrees. This is called the conversion angle. The outstanding question is whether this should be added to or subtracted from the great circle path. This is decided by remembering that great circles always lie to the poleward side of the equivalent rhumb line and therefore, for a generally westerly great circle bearing of 242 in the southern hemisphere the equivalent rhumb line would be 247 degrees.

If you wanted to plot the rhumb line from the beacon to the aircraft as the question asks you would plot the opposite, 067 degrees.

The second part of the question asks the initial magnetic great circle track to fly. We have already established the QDM is 227, 7 degrees of right drift would require a mag heading of 220 degrees.

I'm guessing this is a school question. How did I do?


you did great, Thanks a lot.

[paco]

"This is called the conversion angle. The outstanding question is whether this should be added to or subtracted from the great circle path"

Rum comes from Jamaica - Jamaica is near the Equator....... :)

[SkillsToBurn]

Hello again!

I had 2 new POF questions in my exam that I wasnt too sure of!

Aircraft on Takeoff Run , experiences engine failure, it can maintain its control on Ground with wings level and center line with full rudder deflection ; which calibrated speed will be used as a skill of the pilot:

a)- V1

b)- V2

c) - VMCG

d)- VMU

I guessed C, as you would need to apply rudder to stop you coming off center!

Question of Change is speed of 70Kt to .................. fill i the blank when load factor is increased from 1g to 2.5 G, answer nearest to 2 decimal places.

I did 70xSqaure root of 2.5 = 110.68

If anyone could shed more light on these would be much appreciated
thank you

[Alex Whittingham]

"which calibrated speed will be used as a skill of the pilot" .... what on earth does this mean? Was this in a UK exam?

[GlenQuagmire]

"which calibrated speed will be used as a skill of the pilot" .... what on earth does this mean? Was this in a UK exam?

In the same way that Vmcg cannot be greater than V1, the ICAO English level of the question setter cannot be less than that of the reader. Sadly, in this case, the question speared off the side of the runway and exploded in a shower of vowels.

[emz1234]

NEW ATPL HPL Questions

Whilst taxiing towards the runway, the pilot starts his power checks approaching intersection B2. ATC then clear the aircraft for take off. The pilot replies saying he cannot take off as he isn't prepared for departure as of yet. What kind of behaviour is this?

Unprofessionalism
Over-confidence
Self-disciplined<<<<<<<<<<<<
Anti Authority

An aircraft is approaching an airfield for landing, it has been told to squawk 4292 and has set to the aerodrome QNH of 1002 hPA. The ATC clear the aircraft to land "G-ABCD cleared to land runway 09, surface wind 230 5 knots."

What information is stored in the working memory?

Squawk code<<<<
Local QNH
Wind velocity and vector<<<<
ATC frequency

Question about an aircraft taking off with ice on the wings. The cabin crew knew about the icing due to passengers commenting on it but were resistant to tell the pilots. What kind of error is this?

Ergonomic
Economic
Knowledge based
Social<<<<

with regard to the TEM model what is a systemic “hard” countermeasure that flight crews
used to avoid threats, errors, and undesired aircraft states.
- Read-back of ATC clearances <<<<<<
- ACAS (Airbourne Collision Avoidance System)
- 2 other less convincing options

<<<<< possible answer. If anyone could help me with these that would be most helpful
Thank you
Em

[rudestuff]

The first pilot sounds very self disciplined.

[Piloto Maluco]

"This is called the conversion angle. The outstanding question is whether this should be added to or subtracted from the great circle path"

Rum comes from Jamaica - Jamaica is near the Equator....... :)

Hi Paco

I didn't get your Jamaican mnemotechnics ...

[paco]

Which way to put the rhumb line against a great circle. They make rum in Jamaica, which is near the Equator, so that's where the rhumb line goes. The great circle goes towards the nearest Pole.

[superflanker]

Given the parameters:
TAS 250 KTS
HDG 080o (T)
TRK 090o (T)
GS 290 KTS

What is the value of the crosswind on your heading and which way is it blowing you ?

49 Kts Right.

If i do this with trigonometry I get the correct result, but with CR3 I get 44-45 Kts instead of 49. Can this problem be solved with CR3?

[Capt Pit Bull]

emz1234

Would you be remembering your squawk or would you have dialled it into your transponder?

I hope that answers your question.

[dook]

been told to squawk 4292

Impossible ! 8 is the biggest squawk number with binary octal coding.

[wiggy]

Well the binary octal coding bit behind it is correct AFAIK but that gives you 8 digits : 0-7...

i.e. 7 is the biggest squawk digit.

[Piloto Maluco]

Is it normal that I am struggling with AGK ? This is my last subject but I am looking like this >_<

[eckhard]

Yes, which is why there are a total of 4096 codes available.

(7777 to the base 8 = 4096 to the base 10)

[Rolercoaster79]

Hi all this is my first post here and I hope it is the right topic.
Here is the situation...I am starting to study for ATPL and found that there is Oxford NPA29 1st edition after 6th edition that was before that one.
Does anybody know what is the exact difference between those two? Iam looking to buy paper bundle and just want to make sure that I Will study current sylabus. There is a lot of different used books on ebay but I don’t want to invest in them if there is too big difference.

Any coments and advice?

Thank you all in advance!

[Officer Kite]

AGK is difficult at the start, and for me it was the worst, but it ended up being one of my highest. If you start getting your mind melted over what material the inner workings of a shunt wound generator are made of you are doomed. It's both not required knowledge and will eat up a whole day's worth of study for no benefit whatsoever.

Know where the goods are and what topics are really going to get you the marks.

[keith williams]

Question 1.
In a fully powered flying control system the cockpit controls are connected to the pilot valves. The pilot valves control the flow of pressurised fluid to and from the actuators, thereby exerting very large control forces onto the control surfaces. These forces are so large that the aerodynamic loads cannot be fed back through the system. So there is no aerodynamic feedback, which means that servo tabs cannot be used. The term Servo Valve is just another name for the Pilot Valves. So taking the above into account the only correct answer is Servo Valves.

Question 2.
This one requires a bit of thought. If we enter ground effect at CONSTANT PITCH ATTITUDE the ground will reduce down wash, thereby increasing our ANGLE OF ATTACK. This will in increase cl thereby increasing lift. But the reduced down wash will decrease cdi thereby decreasing drag. But the question specifies CONSTANT ANGLE OF ATTACK. To achieve this we must push the nose down to decrease PITCH ANGLE such that we have CONSTANT ANGLE O F ATTACK. In this situation we we have no increase in Cl, but the reduced down wash will decease the cdi. Taking the above into account the only correct answer is induced drag decreases.

Question 3.
Any worthwhile book dealing with high speed flight will include diagrams showing the way CL and CD vary when accelerating through the transonic speed range. The diagram for CL will show that it increase at speeds just above Mcrit, it then decreases until increasing slightly just before Mcdr, before decreasing again. The overall effect is a decease, but the best option in this question is that CL varies with Mach Number.

[BaronVonBarnstormer]

Ok so came across this today in a Met practice:

"If the QFE, QNH and QFF have the same value,

a)The 1013.25hPa level must at MSL
b)The airport must be at MSL
c)The conditions must be as in the ISA
d)The airport must be at MSL and the conditions must be as in the ISA"

The answer given is b). However I chose d) for the following reason:

We know QFE=QNH=QFF. We also know that a measurement of QFE is used to calculate QNH using the ISA standard lapse rate and QFF is calculated using the actual lapse rate per unit of height. Then it stands to reason that in this condition, if QFE was to vary then QNH would remain equal to QFF. Say for example QFE was then taken at height 'Z' above the airfield, and the lapse rate conditions on the day = LR1 and ISA standard lapse rate = LR2.

QFF = QFE+(LR1*Z), QNH = QFE+(LR2*Z).

If QFF = QNH then QFE+(LR1*Z) = QFE+(LR2*Z)

cancelling for QFE and Z gives LR1 = LR2

hence, Lapse rate conditions on the day = ISA Standard Lapse Rate.

In answer b) I accept that it is not a wrong answer per say, however because QNH and QFF by definition are calculated, then surly d) is the more valid answer?

[Alex Whittingham]

1. If QFE = QFF then the measured pressure at the aerodrome is the same as the actual sea level pressure (near as it can be calculated). If the pressures are the same and the location is the same then the aerodrome must be at sea level.
2. QNH pressure is the sea level pressure calculated back from the observed QFE using ISA lapse rate and the known aerodrome elevation above msl.
3. That's going to come out to the same as QFE as we have already established that the airfield is at sea level, there is no elevation above msl and therefore no difference between QFE and QNH.

So what if the atmosphere was other than ISA? The first statement still stands, it doesn't depend on temperature or lapse rate. The last statement still stands as there is still no elevation above msl. Neither are dependant on ISA lapse rates.

[BaronVonBarnstormer]

1. If QFE = QFF then the measured pressure at the aerodrome is the same as the actual sea level pressure (near as it can be calculated). If the pressures are the same and the location is the same then the aerodrome must be at sea level.
2. QNH pressure is the sea level pressure calculated back from the observed QFE using ISA lapse rate and the known aerodrome elevation above msl.
3. That's going to come out to the same as QFE as we have already established that the airfield is at sea level, there is no elevation above msl and therefore no difference between QFE and QNH.

So what if the atmosphere was other than ISA? The first statement still stands, it doesn't depend on temperature or lapse rate. The last statement still stands as there is still no elevation above msl. Neither are dependant on ISA lapse rates.

Thanks Alex. I think I was trying to "Einstein" it.

[177]

Are SRA and PAR approaches commonly used in Europe?I just finished reading the ICAO Doc 9432 which wasn't updated since 2007. How common are SRA and PAR approaches these days? Are they still practiced anywhere in Europe? Thanks!

[177]

Why there are so many approach lighting systems ?Why not just have only 2 or 3? What's the point of having such and such amount of lights with such and such displacement from each other? What's the PRACTICAL application of each lighting scheme? I mean, do they really help pilots in a different kind of way?

Picture url:
image.ibb.co/fvbhRy/1.jpg

[erikfj]

Hello everyone,

Just stumbled upon the E ( LO ) 2 Chart. The standard parallels for the chart are at 37º and 65º! :confused:
If the maximum latitude range for a Lambert Conformal Conic Chart is 24º ( of which the Standard Parallels should be at no further spread than 16º ) then what kind of corrective calculations must be made for this chart? Or am I missing something?? Can't get my head around it :ugh:

Thanks everyone!

[dook]

I have e-mailed an ex-colleague of mine, known as Mr. Navigator.

He has written a couple of books on the subject and has been teaching this stuff for over forty years. No reply yet - watch this space.

[erikfj]

I have e-mailed an ex-colleague of mine, known as Mr. Navigator.

He has written a couple of books on the subject and has been teaching this stuff for over forty years. No reply yet - watch this space.

Will be! Thank you for the quick response!

[dook]

erikfj,

Here's the response from my colleague:

The standard parallel separation quoted is the ideal separation for maximum accuracy of detail.
There is no correction to apply for the SP’s being other than ideal because the scale depicted is absolutely correct for that chart.

[Alex Whittingham]

That doesn't sound quite right. The quoted maximum SP separation is to ensure that scale expansion and contraction does not exceed 1% (from memory) and the implication must be that, if a larger SP spacing is chosen scale expansion and contraction exceeds 1%. Its nothing to do with 'accuracy of detail'. The scale is not 'absolutely correct', it is approximate, unless the large projection used to generate E(LO)2 is compensated for by having different, more accurate, scales quoted when sections of the projection are taken, more accurate for their latitude with respect to the SP. Still they will not be 'absolutely correct', only approximations.

[selfin]

I just finished reading the ICAO Doc 9432 which wasn't updated since 2007. How common are SRA and PAR approaches these days? Are they still practiced anywhere in Europe?

ICAO Doc 9432 was based on the first edition of the UK Radiotelephony Manual CAP 413 which is now in its 22nd edition.

In UK a PAR talkdown can be done at over 20 military aerodromes but it is unavailable to most civil pilots. Elsewhere in Europe, perhaps in Germany or Sweden, it may be possible at one of the joint civil–military aerodromes.

SRAs can still be done at about two dozen aerodromes in UK although only a handful offer half-milers. Details in AIP.

Why there are so many approach lighting systems? Why not just have only 2 or 3?

There are only about half a dozen different systems and they mainly categorised by the instrument approach procedures served. The choice of system depends on budget constraints, the type/volume of traffic intended, the need for reduced instrument landing minimums, the surrounding topography, and so on. Supplementary lighting such as bars for roll guidance near the threshold will be used for CAT III precision approaches, etc. Some history on the Calvert system is at Calvert Cross Bar Lighting System (http://www.airwaysmuseum.com/Calvert%20cross%20bar%20lighting.htm)

What's the PRACTICAL application of each lighting scheme? I mean, do they really help pilots in a different kind of way?

Visual identification of the intended landing surface during instrument conditions, indication of deviation from the extended runway centreline, roll guidance, and so on. The AVweb article Approach Light Secrets by Jeff Van West, Apr 2014, while intended for a US audience covers much common ground and addresses your questions: https://www.avweb.com/news/features/Approach-Light-Secrets-221926-1.html

For a pertinent technical report from the FAA's Airport Technology Branch see Reduced Approach Lighting Systems (ALS) Configuration Simulation Testing (DOT/FAA/AR-02/81 (http://www.tc.faa.gov/its/worldpac/techrpt/ar02-81.pdf); Gallagher, DW. Jul 2002). Here is the abstract:

The availability of Global Positioning System (GPS) approaches has already increased the number of runways capable of handling Instrument Flight Rule (IFR) approach operations. A major factor in upgrading the instrument capability of these runways is, and will remain, the need for installation of many new approach lighting systems (ALS). Therefore, it has become necessary to re-evaluate the present standard systems to identify possible means by which installation, operation, and maintenance costs can be reduced.

In an effort to reduce the overall length of ALS's, this report describes the methods, using simulation, by which the minimum visual cues with respect to length of an ALS is needed by pilots during an approach at Category I minimums. The current US standard is the 2400-foot-long Medium Intensity Approach Lighting System with Runway Alignment Indicator Lights (MALSR). Subject pilots evaluated ten different length configurations and were given questionnaires for each configuration flown.

The results indicate that shortening the system to a length of 1600 feet was not acceptable. Shortening the system to a length of 1800 or 2000 feet may be conceivable if enhancements to the visual segment portion of the system (i.e., additional steady burning barrettes at 1600,1800, and/or 2000 feet) would be considered. Shortening the system to a length of 2200 feet will only provide minimal reduction in ground area required and result in virtually no benefit in reduced equipment or power requirements.


The standard parallel separation quoted is the ideal separation for maximum accuracy of detail.

That depends on the spatial map extent and choice of distortion measure to be optimised among other factors, see eg first couple of sections in: Savric B, Jenny B. 2016. Automating the selection of standard parallels for conic map projections: Computers and Geosciences, 90, 202–212. (PDF (http://berniejenny.info/pdf/2016_Savric_Jenny_Standard_parallels_for_conic_map_projectio ns.pdf)).

See also p 91 (and footnote 24 re the figure Alex mentions on scale error) in Deetz CH, and Adams OS. 1934. Elements of map projection with applications to map and chart construction (4th ed): U.S. Coast and Geodetic Survey Spec. Pub. 68. (PDF (http://www.michaelbeeson.com/interests/Cartography/References/ElementsOfMapProjection.pdf)):

In general, for equal distribution of scale error, the standard parallels are placed within the area represented at distances from its northern and southern limits each equal to one-sixth of the total meridional distance of the map. It may be advisable in some localities, or for special reasons, to bring them closer together in order to have greater accuracy in the center of the map at the expense of the upper and lower border areas.

[erikfj]

Originally Posted by Deetz and Adams (1934, p 91)
In general, for equal distribution of scale error, the standard parallels are placed within the area represented at distances from its northern and southern limits each equal to one-sixth of the total meridional distance of the map. It may be advisable in some localities, or for special reasons, to bring them closer together in order to have greater accuracy in the center of the map at the expense of the upper and lower border areas.

As stated, It makes perfect sense to me that if needed, greater accuracy can be achieved by selecting two SPs at a range closer to 16º. But why choose two parallels with a change of 28º especially when the chart's ( E (LO) 2 ) latitude coverage ranges only from 52º 20' to 56º 20'? I see no point in selecting those SP and having the chart lie between the Parallel of Origin and the northerly SP where scale contraction error would be almost at its peak and surely above 1%.

I also must admit the math level of 'Savric B, Jenny B. 2016. Automating the selection of standard parallels for conic map projections: Computers and Geosciences, 90, 202–212.' is beyond my current knowledge. I would've loved to post the result to the distortion measure ( 204 Eq(2) ) of the chart though.

[Alex Whittingham]

I think they create a large(ish) mathematical projection then take sections of it for the charts it covers

[selfin]

I agree with Alex. Common standard parallels are used in some regional sets. The publisher in question addresses this in its FAQ (http://ww1.jeppesen.com/main/store/legal/charts/vfr_eu_faq.jsp):

Q. Why does the scale of the VFR+GPS charts differ slightly from 1:500.000?

A. In order to support the European wide standard, a Lambert Conformal Conic Projection with Standard Parallels at N37° and N65° is used for production of all VFR+GPS charts. This enables the seamless combination of all charts and the use for continuous flight planning. A scale distortion is a consequence of the projection and varies with the distance from the Standard Parallels. The true scale is depicted on every VFR+GPS chart on the Cover panel (e.g. EB/EH: True Scale at N51° - 1:515.000) and near the Scale Bar. In addition, a factor is given to multiply measured distances with every parallel (e.g. 1.03).

greater accuracy can be achieved by selecting two SPs at a range closer to 16 [deg]

While this may be true for the chart in question it obviously depends on the actual latitude range to be charted.

I also must admit the math level ... is beyond my current knowledge.

Ignore it. The pertinent information was in the text and was adequately summarised in Deetz and Adams.

In practice because the scale on these projections is almost constant it is easier to assess distances by referring to a meridian graduated in minutes of latitude.

[UWGGair]

True Altitude Calculation

Hello

Could you please provide some assistance with the following question?

Altimeter reading 4500 FT, OAT 20°C, calculate the true altitude... has it got anything to do with pressure or density altitudes? :bored:

[erikfj]

I agree with Alex. Common standard parallels are used in some regional sets. The publisher in question addresses this in its FAQ

Quote:
Q. Why does the scale of the VFR+GPS charts differ slightly from 1:500.000?

A. In order to support the European wide standard, a Lambert Conformal Conic Projection with Standard Parallels at N37° and N65° is used for production of all VFR+GPS charts. This enables the seamless combination of all charts and the use for continuous flight planning. A scale distortion is a consequence of the projection and varies with the distance from the Standard Parallels. The true scale is depicted on every VFR+GPS chart on the Cover panel (e.g. EB/EH: True Scale at N51° - 1:515.000) and near the Scale Bar. In addition, a factor is given to multiply measured distances with every parallel (e.g. 1.03).

Awesome! thank you very much! This clears all the doubts.

@dook , @Alex Whittingham and @selfin thank you very much for your help. I appreciate all the help and input you gave.

[Specaircrew]

True Altitude Calculation

Hello

Could you please provide some assistance with the following question?

Altimeter reading 4500 FT, OAT 20°C, calculate the true altitude... has it got anything to do with pressure or density altitudes? :bored:

They've given you the OAT at 4500ft which means it's an ISA deviation question. You know what the ISA temp is at sea Level (15C) so you work out the ISA deviation then use the formula 4 x Indicated Alt in 1000's of feet x the ISA Deviation to calculate the height error.

[Platypus]

Hello everyone,

I am really struggling with two types of M&B questions, someone can help with a scheme?

First one:

"The loaded weight of an aeroplane is 100.000 kg.
The CG of the loaded aeroplane is at 20% MAC = Sta. 15,5 m.
The CG should be shifted to Sta. 16 m by moving cargo from the fwd. hold (sta. 10m) to the aft. hold (sta. 25m).
Initially, fwd. cargo load is 5.000kg and the aft. cargo load is 3.000kg.
How much cargo load is in the aft hold after the load shift to obtain the new CG?"

Second one:

"The loaded weight of an aeroplane is 13.000kg.
The CG of the loaded aeroplane is at Sta. 105,5in.

The aft. CG limit is at Sta. 102in.

How many seat rows (seat pitch 33in) must four passengers (75kg each) move forward from the last row (Sta. 224in), to bring the CG at least to the aft. limit?"

[keith williams]

To solve any problem we need to begin by forming a clear picture of the situation.

In these two questions we are attempting to achieve a certain moment change by moving part of the load.

The moment change which we are trying to achieve is equal to the total mass multiplied by the distance we want to move the CofG. We could express this as M x S where M is the total mass and S is the required CofG shift.

We will achieve this by shifting a small part of the mass a certain distance. We could express this moment change which we are going to cause as m x s where m is the mass we move and s is the distance we shift it.

If we do the job properly the moment change we are trying to achieve (M xS) will be equal the moment change we cause by shifting the smaller mass (m x s)

So we will have M x S equals m x s

In the first question we are trying to calculate the mass to be moved, (s), so we rearrange the equation to give

(M x S) / m equals s (sorry my tablet does not appear to have an equals sign)

M is 100000, S is (16 - 15.5) which is 0.5, s is (25 - 10) which is 15

Inserting these number into our equation gives (100000 x 0.5) / 15 equals the mass to be moved to the aft hold.

Adding this to the initial mass of 3000 kg already in the hold gives us the answer.

The second question can be solved in a similar way, but this time we are trying to find the distance we need to move the 300 kg of the 4 passengers. Dividing this by the seat pitch will then give us the number of rows.

[ersin737]

I think EASA published ECQB 6 new ATPL Question bank. and its already has been using by EASA caa. probably we will see more questions like that.

[77Whiskey]

Hi everyone, A newbie here in the search of some help. I can't figure out how to solve this question of altimetry.
Question :
An aircraft takes off from A (elevation 600ft) aerodrome pressure 1008mbs. The altimeter on QNH reads 630 ft on ground has to clear a 7210 ft high hill midway by of margin 1500 ft before landing at B (elevation 330 ft) QFE = 1005 mbs. If A/C maintain QNH of A throughout, find

Minimun altimeter reading to clear the hill.
Aircraft altimeter reading on landing.

[WASALOADIE]

If using 1Mb as 30ft (rounded up from 27ft) Then the QNH allowing for the over read of 30ft (630ft Alt) at A should be 1029. In theory, using 1029 and maintaining 8710 ft Altitude should give you the 1500ft clearance.
However! The pressure at B is lower (QNH + 11Mb = 330ft elev = 1016) Therefore, flying from a high pressure area to a lower pressure area, the adage "High to low, beware below" means that the altimeter reading a constant altitude but the aircraft would in fact descend in relation to the MSL. So you would need to fly higher. My suggestion would be the difference between 1029Mb and 1016 Mb ( 13Mb = 390ft) minimum (Altitude of 9100ft).
If you require more accurate, then use 27ft per Mb

[erikfj]

Hi everyone!

Quick and probably dumb question regarding ATS Routes: What does ATS regional routes actually mean ?
I cannot find, in either the Jeppesen or anywhere else, any explanation beyond that they are used by international traffic. Does it mean that a "ULXXX" RNAV airway for example will not cross borders between states (countries)? Why are certain Juliet airways that do cross borders then?

[KT1988]

Hi everyone!

Does anyone have any idea if the amount of questions about anatomy of the eye, ear etc. in Human Performance & Limitations have been reduced or something? Do Aviation Exam have outdated questions or just some question banks are lacking important questions?

BTW of course its important to learn stuff from the bristol ground school and the books etc. But where are you supposed to get info for example that dengue fever is transmitted by mosquitoes active by day and not by those active at night? The question bank say such a question exist for real on the exam. And what relevance does knowing it have to flying? But you still have to know the answer to get a nice result.

[paco]

Dengue fever is in our notes......

[asmith474]

There are PPL questions but I just dont understand how to figure them out. Any feedback is appreciated.

1) The aircraft takes off from the airport elevation 500 ft MSL and rises vertically at a speed of 500 ft / min. Its average cruising speed of 100 knots, and the pressure = 1013.2 hPa QNH. How far from the airport FL 80 is reached?

Select one:
a. 21 NM; b. 25 NM; c. 18 NM; d. 12.5 NM;

2)Compass deviation = -3° , Variation = 2°E, Compass heading = 127°. Magnetic track and true track values are respectively:

Select one:
a. 126°, 128°; b. 126°, 124°; c. 124°, 124°; d. 124°, 126°; .

3)An aircraft flying in a windless conditions with the heading 320 crosses the 195 radial from the VOR JED. The aircraft will be:

a. To the east of the VOR-a JED.
b. Over the VOR-em JED.
c. To the west of the VOR-a JED.
d. None of the above

[KT1988]

@paco: Dengue fever is mentioned in the theory and that mosquitoes spread it, but I fail to remember a statement that mosquitoes active by day but not those active at night spread it. But maybe I have not read the less relevant to flying stuff good enough (its hard to get everything perfectly among stuff that have little relevance to flying).

@asmith474: The QNH is exactly the same as ISA value used for flight levels 1013.2 hPa so the only thing you do is 8000 ft - 500 ft = 7500 ft. Then 7500 ft / 500 ft/min = 15 min. Then just 100/60 = 1,666667 and 1,66667 x 15 = 25 NM or just 0,25x100 = 25 NM since 15 min is 0,25 hour.

In question nr.2 are you sure they did not say anything about a drift angle? or wind direction and velocity?

As for question 3 just look at your CRP-5 (or get one you will need it in the future if you plan to do the ATPL) and imagine the VOR in the middle. Then radial 195 is going southwest from the VOR (more south than west) and aircraft flying 320 heading (magnetic since we do not know anything about variation) is going northwest. So its crossing a radial that is to the west of the VOR JED and its going in a northwesterly direction. So it has to be in the west of the VOR JED (but of course not exactly to the west, its southwest of the VOR JED in the moment of the crossing of the radial 195 so a tricky question giver could imagine that to the west means on radial 270 so then none of the above would be correct, thats why question banks help before the exam to solve different tricks from the question creators).

[RHSandLovingIt]

Q2 is just your typical Deviation/Variation question...

You need to remember three four things:

1. Compass <-deviation-> Magnetic <-variation-> True
2. "Deviation West, Compass Best, Deviation East, Compass Least"
3. "Variation West, Magnetic Best, Variation East, Magnetic Least"
4. West = Negative, East = Positive

So, you have the compass heading 127 and you need magnetic track... deviation is -3... negative = west, so 3W... "deviation west, compass best", so we need to subtract from compass heading: 127-3 = 124 magnetic heading.
Now for magnetic to true... variation is 2E... "variation east, magnetic least", so we need to add to magnetic to find true. 124+2 = 126 true.

Answer should be D: 124°, 126°; (assuming nil wind... as these are technically "headings", so as KT1988 pointed out, if they mention drift angles/wind etc, then you have to factor those in to get "tracks")

[KT1988]

Have anyone done recently the exam in Radio Navigation? Do you know approximately how many of the 66 questions were about the PBN? Because Aviation Exam and other training databases got very few PBN questions and I heard they can be quite many on the exam so when training the questions in the data banks I do not know if I really know the whole subject or if I just remember the questions.

[paco]

KT - just in case they ask you which mosquito :)

Dengue is another fever spread by infected mosquitoes that are active by day – usually the Aedes Aegypti.

With regard to PBN, probably about 5 in the RNAV exam. In the separate exam that you will have to take if you don't do it now, it's 25,

[KT1988]

@paco: Thanks, I did already the human performance exam, they did not ask at all about the mosquitoes on the real exam and most of the questions made sense, some were tricky but the result was 90 % so I can live with it (since I used most time learning General Navigation for that session).

So its about 5 questions of the 66 about PBN on the Radio Navigation exam? I am going to do the exam on 2nd April when next session start (If I get done with everything including June session, I got promised I will be done with all my licenses this year so I am learning everyday 14/7 rest is eating, sleeping or writing posts) and heard from a colleague who were not too lucky on the radio navigation exam that he got like 14-15 PBN questions from the 66 and that he got still RNAV (area navigation questions they were not replaced by PBN like Bristol ground school online suggested).

But maybe its the Polish exam where they added PBN questions but did not remove the previous area navigation that UK CAA removed according to Bristol. So I learned both and practice with both kind of questions. But using more time on one subject means less time used on the instruments, operational procedures and air law that I also plan to pass in April. So thats why I asked if anyone know if PBN replace the RNAV part of Radio navigation exam or if its just an addition using some of the RNAV question slots on the exam but not replacing it totally.

PS. From what I heard since September 2018 its not possible to do radio navigation exam without PBN so everyone doing the exam now will get it done with PBN.

[paco]

14-15 sounds excessive

[asmith474]

Some PPL navigation questions that I dont understand

1)DME station is situated at an altitude of 1000 ft AMSL. QNH = 1013.25 hPa. The aircraft flies at FL370, 15 NM DME from the station. DME readout will be:
a.16nm b.15nm c18nm d.37nm

2)Variation = 3° E, Magnetic track = 188 °, Compass heading = 190 °. True track and Compass deviation values are respectively:
a. 194°, +4. b. 189°, -3. c. 185°, -2. d. 191°, -2.

3)In windless conditions at 0830 UTC Pilot read radial = 315 degrees. Continuing flight with true track heading = 180 degrees at 0840 UTC he reads radial = 270 degrees. Assuming his ground speed is 240 kt, calculate distance to NDB at 0840 UTC:
a. 32 NM. b. 48 NM. c. 38 NM. d. 40 NM.

[77Whiskey]

1st : Station ALT 1000ft AMSL, So we'll subtract station height from 37,000(FL370) so 36,0000 (which converters to nearly 6NM) and the distance is 15 NM. We need to find the slant range. So formula of slant range is (S.R)^2 = (Height 1)^2+(Height 2)^2
(S.R)^2 = (15)^2 + (6)^2
(S.R)^2 = 225 + 36
(S.R)^2 = 261
S.R = √261
S.R = 16.1 NM, So the answer is A

2nd : You have the compass and magnetic heading 190° and 188° respectively. Remember Variation east magnetic least (-) so in order to get true track you need to go east(-) because variation is 3°E from the magnetic track. Therefore 188°-3° = 185°(T). The deviation was east(-) of magnetic by 2° So, -2°

3rd : I think it's a question of 1 in 60 rule but not sure

[asmith474]

4) The weight and balance sheet of an aircraft gives the following data: nose wheel weight 1000kg, left and right wheels 5000 kg each. The distance between the nose wheel and the main ones is 10 m. How many meters from the main landing wheels is located the centre of gravity?

a. 0.81m b.0.75m c. 0.91m d 9.1m

5) In windless conditions at 0830 UTC Pilot read radial = 315 degrees. Continuing flight with true track heading = 180 degrees at 0840 UTC he reads radial = 270 degrees. Assuming his ground speed is 240 kt, calculate distance to NDB at 0840 UTC:

a. 32NM b. 48NM c. 38NM d. 40NM

(Exams is in 2 days, so any help is appreciated!)

[paco]

FL 370 for a PPL question?
An 11000 kg aircraft?

[asmith474]

1st : Station ALT 1000ft AMSL, So we'll subtract station height from 37,000(FL370) so 36,0000 (which converters to nearly 6NM) and the distance is 15 NM. We need to find the slant range. So formula of slant range is (S.R)^2 = (Height 1)^2+(Height 2)^2
(S.R)^2 = (15)^2 + (6)^2
(S.R)^2 = 225 + 36
(S.R)^2 = 261
S.R = √261
S.R = 16.1 NM, So the answer is A

2nd : You have the compass and magnetic heading 190° and 188° respectively. Remember Variation east magnetic least (-) so in order to get true track you need to go east(-) because variation is 3°E from the magnetic track. Therefore 188°-3° = 185°(T). The deviation was east(-) of magnetic by 2° So, -2°

3rd : I think it's a question of 1 in 60 rule but not sure

The 1st answer was correct but the 2nd was wrong according to the question bank results

[oggers]

15 NM DME from the station. DME readout will be:
a.16nm b.15nm c18nm d.37nm

:rolleyes: If the aircraft is at 15 DME (which the question states it is) the only possible correct answer is b. 15 DME! Perhaps find a better question bank.