ATPL theory questions
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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.
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.
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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
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
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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
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
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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
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
Last edited by Ronaldsway Radar; 20th Mar 2018 at 17:40. Reason: Clarification
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Ratio Question
Hey everyone
I'm a little stuck with this
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%?
I'm a little stuck with this
Any help would be appreciatedNeed 2.6% climb gradient, currently have 2.8% at 110000KG. What will the new
MTOW to have 2.6%?
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How much is that increase? 2.8/2.6 = 1.077 --> 110000 * 1.077 = 118470Kg aprox.
Last edited by superflanker; 22nd Mar 2018 at 07:15.
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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.
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.
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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?
T + W sine theta = D
or T = D - W sine theta.
substituting,
T = 8500 - (11000*0.5) = 3000 Kg. How did I do?
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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!
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!
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hey folks
Can someone help me understand how to solve problems like this? Im kinda stuck
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 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
Can someone help me understand how to solve problems like this? Im kinda stuck
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
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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?
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?
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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?
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.
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"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.......
Rum comes from Jamaica - Jamaica is near the Equator.......
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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
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
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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.
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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?
What information is stored in the working memory?
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
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
What information is stored in the working memory?
- Squawk code<<<<
- Local QNH
- Wind velocity and vector<<<<
- ATC frequency
- Ergonomic
- Economic
- Knowledge based
- Social<<<<
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
Last edited by emz1234; 12th Apr 2018 at 07:30.