ATPL theory questions
Joined: Jun 2017
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From: Madrid
How much is that increase? 2.8/2.6 = 1.077 --> 110000 * 1.077 = 118470Kg aprox.
Last edited by superflanker; 22nd March 2018 at 07:15.
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From: Johannesburg, South Africa
Joined: Aug 2010
Posts: 23
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From: Aberdeen
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.
Joined: May 1999
Posts: 1,846
Likes: 4
From: Bristol, England
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?
Joined: Aug 2010
Posts: 23
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From: Aberdeen
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!
Joined: Apr 2018
Posts: 3
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From: NORWAY
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
Joined: May 1999
Posts: 1,846
Likes: 4
From: Bristol, England
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?
Joined: Apr 2018
Posts: 3
Likes: 0
From: NORWAY
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.
Joined: Nov 2000
Posts: 4,330
Likes: 74
From: White Waltham, Prestwick & Calgary
"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.......
Joined: Aug 2010
Posts: 23
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From: Aberdeen
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
Joined: May 2009
Posts: 137
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From: london
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.
Joined: Apr 2018
Posts: 1
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From: Edinburgh
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 April 2018 at 07:30.
Joined: Jun 2012
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From: Earth
Joined: Nov 2000
Posts: 4,330
Likes: 74
From: White Waltham, Prestwick & Calgary
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.
Joined: Jun 2017
Posts: 153
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From: Madrid
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?
