The first pilot sounds very self disciplined.

Hi Paco

I didn't get your Jamaican mnemotechnics ...

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.

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?

emz1234

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

I hope that answers your question.

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

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

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!

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.

answers to new questions
 

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.

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?

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.

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!

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!

erikfj,

Here's the response from my colleague:

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.

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.

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

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/...-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; Gallagher, DW. Jul 2002). Here is the abstract:


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).

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):

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.

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

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

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

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.

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:

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.

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.

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?"

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.

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.

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

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?

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.

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

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

@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).

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")