# ATPL theory questions

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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.

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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?

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?

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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!

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!

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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.

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.

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.

*Last edited by keith williams; 30th Jun 2018 at 18:32.*

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Ok so came across this today in a Met practice:

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

hence,

In answer b) I accept that it is not a

*"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"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?Join Date: May 1999

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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.

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.

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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.

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## 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!

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## 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?

image.ibb.co/fvbhRy/1.jpg

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

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Hello everyone,

Just stumbled upon the

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

Thanks everyone!

Just stumbled upon the

**E ( LO ) 2**Chart. The standard parallels for the chart are at**37º**and**65º**!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

Thanks everyone!

*Last edited by erikfj; 16th Aug 2018 at 06:03.*

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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.

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.

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