ATPL theory questions
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You might get INS questions in Nav - and remember that the Nav exam has been traditionally a dustbin for all sorts of stuff that doesn't fit into other subjects - we tell our guys to expect the unexpected! 
Sometimes you might get a question that simply asks you to move figures around a formula.
One tip if you get a mention of an INS in Nav is to realise they are really talking about great circles.
Sometimes you might get a question that simply asks you to move figures around a formula.One tip if you get a mention of an INS in Nav is to realise they are really talking about great circles.
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Help ?
Can anyone please solve this question for me
Flying at FL270, an AWR weather return at range 40nm is identifiable when the centre of the beam is tilted between +1/-1deg & -6deg. What is the height of the base & top of the rain bearing cloud?
Flying at FL270, an AWR weather return at range 40nm is identifiable when the centre of the beam is tilted between +1/-1deg & -6deg. What is the height of the base & top of the rain bearing cloud?
Are you sure that you are quoting the correct data?
This type of question usually includes a tilt up angle, a tilt down angle and a beam width. You can then use the range and the 1 in 60 rule to work out the solution.
But your quoted figures of between
do not make any sense.
This type of question usually includes a tilt up angle, a tilt down angle and a beam width. You can then use the range and the 1 in 60 rule to work out the solution.
But your quoted figures of between
+1/-1deg & -6deg
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Hi, i'm a bit confused as on my CQB some questions of Air Law are out of date i suppose:
1) The VMC minima for an airspace classified as "B" above 10 000 feet MSL are: 8 KM CLEAR OF CLOUDS
Seems that now ICAO regulations are 8km 1.5 vis and 1000ft as according to jar ops..?
2)The protection areas associated with instrument approach procedures are determined with the assumption that turns are performed at a bank angle of:
SIGNED CORRECT:
25° or the bank angle giving a 3°/s turn rate, whichever is lower, for departure, approach or missed approach instrument procedures, as well as circling-to-land (with or without prescribed flight tracks).
What about this one??should be 20° for circling and 15° for missed app?
3)
According to JAR-FCL, Class 2 medical certificate for private pilots will be valid for SIGNED CORRECT:
60 months until age of 30, 24 months until age of 50, 12 months until age of 65 and 6 months thereafter
Shouldn't it be 60 until 40, 24 until 50 and 12 thereafter..?
4)
Runway edge lights excepted in the case of a displaced threshold shall be: SIGNED CORRECT:
Fixed variable white.
Isn't fixed variable white or yellow?
Thank you
1) The VMC minima for an airspace classified as "B" above 10 000 feet MSL are: 8 KM CLEAR OF CLOUDS
Seems that now ICAO regulations are 8km 1.5 vis and 1000ft as according to jar ops..?
2)The protection areas associated with instrument approach procedures are determined with the assumption that turns are performed at a bank angle of:
SIGNED CORRECT:
25° or the bank angle giving a 3°/s turn rate, whichever is lower, for departure, approach or missed approach instrument procedures, as well as circling-to-land (with or without prescribed flight tracks).
What about this one??should be 20° for circling and 15° for missed app?
3)
According to JAR-FCL, Class 2 medical certificate for private pilots will be valid for SIGNED CORRECT:
60 months until age of 30, 24 months until age of 50, 12 months until age of 65 and 6 months thereafter
Shouldn't it be 60 until 40, 24 until 50 and 12 thereafter..?
4)
Runway edge lights excepted in the case of a displaced threshold shall be: SIGNED CORRECT:
Fixed variable white.
Isn't fixed variable white or yellow?
Thank you
Last edited by MatteoLIPH; 28th Jun 2013 at 10:30.
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The Oxford ATPL book of Principles of Flight has references that I need some help with.
For e.g. On Page 174; Chapter-Stalling
It talks about EASA 25.103(b) on a few occasions.
Now, where can I look up this regulation?
For e.g. On Page 174; Chapter-Stalling
It talks about EASA 25.103(b) on a few occasions.
Now, where can I look up this regulation?
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It talks about EASA 25.103(b) on a few occasions.
Now, where can I look up this regulation?
Now, where can I look up this regulation?
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ICAO Annex III - Meteorological Service for International Air Navigation, Edition 15, Appendix V (Page 107)
TX25/13Z TN09/05Z
TX05/12Z TNM02/03Z
Maximum temperature 25°C at 1300Z Minimum temerature 09°C at 0500Z
Maximum temperature 05°C at 1200Z Minimum temerature -02°C at 0300Z
TX25/13Z TN09/05Z
TX05/12Z TNM02/03Z
Maximum temperature 25°C at 1300Z Minimum temerature 09°C at 0500Z
Maximum temperature 05°C at 1200Z Minimum temerature -02°C at 0300Z
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ATP Nav Gen question help
Hi guys
I'm currently doing self-studiesfor the ATP preparation and need some helps from anyone who can give me hints to solve out this type of question:
An aircraft leaves position S01 20 W012 15 on a track of 360 degrees true for 01:06 mins.The heading is then altered to make good track of 090 degrees true for 03:12 mins to eestination.If the TAS is 320 kts and the wind is calm throughout the flight,What is the lat & long of the of the destination?
I just need the methodology to lead on how to solve the question
Many thanks in advance
NEXT TIME please post in the correct thread
HWB
I'm currently doing self-studiesfor the ATP preparation and need some helps from anyone who can give me hints to solve out this type of question:
An aircraft leaves position S01 20 W012 15 on a track of 360 degrees true for 01:06 mins.The heading is then altered to make good track of 090 degrees true for 03:12 mins to eestination.If the TAS is 320 kts and the wind is calm throughout the flight,What is the lat & long of the of the destination?
I just need the methodology to lead on how to solve the question
Many thanks in advance
NEXT TIME please post in the correct thread
HWB
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320 kts for 1.1 hours is 352 nm. That's 352 minutes or 5 degrees 52 mins so you end up at the end of the Northern sector at N4 32, then turn right. The problem is now that 1 minute no longer represents 1 nm.......
3 hrs 12 mins at 320 kts is 1024 nm.
See if you can get the rest? Hint: you need to look at departure
3 hrs 12 mins at 320 kts is 1024 nm.
See if you can get the rest? Hint: you need to look at departure
Last edited by paco; 4th Jul 2013 at 14:12.
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I take it that you are not self-studying for an EASA licence........because you can't!
The studying must be done through an Approved Training Organisation (ATO).
However, I've written out a painstaking, line by line solution below, for you to check your answer against.
As previously stated, 5°52' due north from S01°20' W012°15' lands you at N04°32' W012°15'.
To calculate the change of longitude from this position you need to use the "departure" formula:
Departure (NM) = Change of Longitude (mins) x Cosine of Latitude
The Departure (or distance) = 320 kts x 3.2 hours = 1,024 NM
1,024 NM = Change of Longitude (mins) x Cosine 04°32'
1,024 NM = Change of Longitude (mins) x 0.997 (Transpose formula)
Change of Longitude (mins) = 1,024 / 0.997
Change of Longitude (mins) = 1,027'
Change of Longitude = 1,027'/60' = 17.12°
Change of Longitude = 17°07'
To find out the Easterly longitude, subtract 012°15' from 17°07'
17°07' - 012°15' = E004°52'
Final Position = N04°32' E004°52'
Hope that helps.
The studying must be done through an Approved Training Organisation (ATO).
However, I've written out a painstaking, line by line solution below, for you to check your answer against.
As previously stated, 5°52' due north from S01°20' W012°15' lands you at N04°32' W012°15'.
To calculate the change of longitude from this position you need to use the "departure" formula:
Departure (NM) = Change of Longitude (mins) x Cosine of Latitude
The Departure (or distance) = 320 kts x 3.2 hours = 1,024 NM
1,024 NM = Change of Longitude (mins) x Cosine 04°32'
1,024 NM = Change of Longitude (mins) x 0.997 (Transpose formula)
Change of Longitude (mins) = 1,024 / 0.997
Change of Longitude (mins) = 1,027'
Change of Longitude = 1,027'/60' = 17.12°
Change of Longitude = 17°07'
To find out the Easterly longitude, subtract 012°15' from 17°07'
17°07' - 012°15' = E004°52'
Final Position = N04°32' E004°52'
Hope that helps.
Last edited by 2close; 6th Jul 2013 at 00:34.
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Hey guys,
Can anyone send me some examples with the answers regarding Drift Down, Im doing Performance second attempt and there were 2 questions on drift down but I havent seen any in the banks (using easaatp.com bristol and oxford). The driftdown questions were calculations, not just memorization.
I'm also finding different answers for calculating speeds from holding speeds and stall speeds, for prop and jet, any ideas?
Thanks
Can anyone send me some examples with the answers regarding Drift Down, Im doing Performance second attempt and there were 2 questions on drift down but I havent seen any in the banks (using easaatp.com bristol and oxford). The driftdown questions were calculations, not just memorization.
I'm also finding different answers for calculating speeds from holding speeds and stall speeds, for prop and jet, any ideas?
Thanks
Last edited by squall1984; 7th Jul 2013 at 23:35.
Given the following information determine the greatest distance from an obstacle, at the failure of one engine would cause an MRJT1 aircraft to clear the obstacle by the statutory minimum during the drift-down, the fuel used during the drift-down, and time taken to reach the obstacle.
Cruise Pressure Altitude 34,000 ft.
Gross Weight at Engine Failure 54,000 kg.
Ambient Temperature ISA +15°C.
Wind Component 40 kt tailwind.
Engine and Wing Anti-icing System on.
Air Conditioning System on throughout the drift-down.
Obstacle Pressure Altitude 14,000 ft.
One engine inoperative.
1. Select the appropriate graph for the given cruise altitude. For 34000 ft this is figure 4.25, which covers altitude between 33000 ft and 35000 ft.
2. Calculate the height required to clear the obstacle during the descent. The statutory minimum clearance during drift down is 2000 ft, so the required attitude at the obstacle is 14,000 ft + 2,000 ft = 16,000 ft.
3. Using the data in the table at the top left corner of the page adjust the gross weight at engine failure to reflect the conditions of the anti-icing and the air conditioning system. For the stated condition of engine and wing anti-icing systems on and air conditioning system on, the corrections are 54,000 kg + 5,650 kg (for anti-icing)= 59,650 kg. Note that an adjustment is required for air conditioning only when it is switched off, so no adjustment is required in this question.
4. Using the graph at the top right cornet of the page adjust the corrected gross weight at engine failure to reflect the temperature deviation. To do this enter the left edge of the sub-graph at 59,650 kg, then move parallel to the sloping lines to a point vertically above ISA +15°C. From this point move horizontally to the right edge of the sub-graph and read off equivalent gross weight at engine failure on the right vertical axis 61,500 kg.
5. Enter the right edge of the main graph and interpolate between 60000 kg and 65000 kg to locate the 61500 kg point. From this point draw a curve to the left interpolating between the printed curved lines on the graph. In doing this take care to take account of the fact that the curved lines converge as they move to the left.
6. Enter the left edge of the graph at 16,000 ft and draw a horizontal line to the left to intercept the curve drawn in stage 5 above.
7. At the intersection point interpolate between the dotted curved lines to estimate the fuel used during the drift-down. In this question the fuel used is approximately 1000 kg.
8. From the same intersection point draw a line vertically down to the bottom of the gridded area and read off the time required to drift-down. In this question it is approximately 29 minutes.
9. Extend the vertical line down to the reference line in the lower gridded area. From this point draw a line parallel to the sloping lines. From the existing 40 kt tailwind condition marked at the left edge of the graph, draw a horizontal line to the right to intercept the sloping line drawn previously. From the point of intersection of these lines drop vertically down to the bottom of the graph and read of the ground distance of approximately 175 nm.
If you are happy with the one above please try this one (something very like it appeared in a recent EASA ATPL exam)
Given the following information determine the greatest distance from an obstacle, at the failure of one engine would cause an MRJT1 aircraft to clear the obstacle by the statutory minimum during the drift-down, the fuel used during the drift-down, and time taken to reach the obstacle.
Cruise Pressure Altitude 37,000 ft.
Gross Weight at Engine Failure 44,000 kg.
Ambient Temperature ISA -15°C.
Wind Component 30 kt headwind.
Engine Anti-icing System on.
Wing anti-icing off.
Air Conditioning System on throughout the drift-down.
Obstacle Pressure Altitude 23,000 ft.
One engine inoperative.
Answers:
Ground distance 122 nm.
Time = 24 minutes.
Fuel used = 700 kg.
Cruise Pressure Altitude 34,000 ft.
Gross Weight at Engine Failure 54,000 kg.
Ambient Temperature ISA +15°C.
Wind Component 40 kt tailwind.
Engine and Wing Anti-icing System on.
Air Conditioning System on throughout the drift-down.
Obstacle Pressure Altitude 14,000 ft.
One engine inoperative.
1. Select the appropriate graph for the given cruise altitude. For 34000 ft this is figure 4.25, which covers altitude between 33000 ft and 35000 ft.
2. Calculate the height required to clear the obstacle during the descent. The statutory minimum clearance during drift down is 2000 ft, so the required attitude at the obstacle is 14,000 ft + 2,000 ft = 16,000 ft.
3. Using the data in the table at the top left corner of the page adjust the gross weight at engine failure to reflect the conditions of the anti-icing and the air conditioning system. For the stated condition of engine and wing anti-icing systems on and air conditioning system on, the corrections are 54,000 kg + 5,650 kg (for anti-icing)= 59,650 kg. Note that an adjustment is required for air conditioning only when it is switched off, so no adjustment is required in this question.
4. Using the graph at the top right cornet of the page adjust the corrected gross weight at engine failure to reflect the temperature deviation. To do this enter the left edge of the sub-graph at 59,650 kg, then move parallel to the sloping lines to a point vertically above ISA +15°C. From this point move horizontally to the right edge of the sub-graph and read off equivalent gross weight at engine failure on the right vertical axis 61,500 kg.
5. Enter the right edge of the main graph and interpolate between 60000 kg and 65000 kg to locate the 61500 kg point. From this point draw a curve to the left interpolating between the printed curved lines on the graph. In doing this take care to take account of the fact that the curved lines converge as they move to the left.
6. Enter the left edge of the graph at 16,000 ft and draw a horizontal line to the left to intercept the curve drawn in stage 5 above.
7. At the intersection point interpolate between the dotted curved lines to estimate the fuel used during the drift-down. In this question the fuel used is approximately 1000 kg.
8. From the same intersection point draw a line vertically down to the bottom of the gridded area and read off the time required to drift-down. In this question it is approximately 29 minutes.
9. Extend the vertical line down to the reference line in the lower gridded area. From this point draw a line parallel to the sloping lines. From the existing 40 kt tailwind condition marked at the left edge of the graph, draw a horizontal line to the right to intercept the sloping line drawn previously. From the point of intersection of these lines drop vertically down to the bottom of the graph and read of the ground distance of approximately 175 nm.
If you are happy with the one above please try this one (something very like it appeared in a recent EASA ATPL exam)
Given the following information determine the greatest distance from an obstacle, at the failure of one engine would cause an MRJT1 aircraft to clear the obstacle by the statutory minimum during the drift-down, the fuel used during the drift-down, and time taken to reach the obstacle.
Cruise Pressure Altitude 37,000 ft.
Gross Weight at Engine Failure 44,000 kg.
Ambient Temperature ISA -15°C.
Wind Component 30 kt headwind.
Engine Anti-icing System on.
Wing anti-icing off.
Air Conditioning System on throughout the drift-down.
Obstacle Pressure Altitude 23,000 ft.
One engine inoperative.
Answers:
Ground distance 122 nm.
Time = 24 minutes.
Fuel used = 700 kg.
Last edited by keith williams; 10th Jul 2013 at 12:32.
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Here's a question few days ago i found out the answer but forgot the methodology that led me to this answer:
The scale at the equator on a Mercator chart is 1cm to 5nm.How many nm to the inch are there at S 44 00?
The scale at the equator on a Mercator chart is 1cm to 5nm.How many nm to the inch are there at S 44 00?