# ATPL theory questions

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**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.

There is no correction to apply for the SP’s being other than ideal because the scale depicted is absolutely correct for that chart.

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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.

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Originally Posted by

**177**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?

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.

Originally Posted by

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

Originally Posted by

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

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:

Originally Posted by

**DOT/FAA/AR-02/81**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.

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.

Originally Posted by

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

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):

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.

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**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.

**28ş**especially when the chart's ( E (LO) 2 ) latitude coverage ranges only from 52ş 20' to 56ş 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.

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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.

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).

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]

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

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.

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**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?

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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).

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).

@

**dook ,**@

**Alex Whittingham**and

**@selfin**thank you very much for your help. I appreciate all the help and input you gave.

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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?"

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.

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.

*Last edited by keith williams; 7th Oct 2018 at 21:05.*

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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 1008

An aircraft takes off from A (elevation 600ft) aerodrome pressure 1008

**mbs**

**. 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.**

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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

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

*Last edited by WASALOADIE; 13th Nov 2018 at 14:51. Reason: Addition of text*

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Hi everyone!

Quick and probably dumb question regarding ATS Routes: What does

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?

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?

*Last edited by erikfj; 14th Nov 2018 at 23:47.*

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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.

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.

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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

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

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@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).

@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).

*Last edited by KT1988; 14th Mar 2019 at 15:06.*

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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")

You need to remember

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")