Ground School Exam Questions & Question Banks
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Yeah they don't call me NC for nothing you know 
Funny you say that about the CRP, Clowns. Most people in my ATPL class, myself included, much preferred a calculator to the WW. I think the WW is an excellent piece of kit, but there are so many figures and words on it I just wanted that extra bit of security and I know my calculator pretty well funnily enough.
Funny you say that about the CRP, Clowns. Most people in my ATPL class, myself included, much preferred a calculator to the WW. I think the WW is an excellent piece of kit, but there are so many figures and words on it I just wanted that extra bit of security and I know my calculator pretty well funnily enough.
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Hi,
first you have to do pressure correction with: 27ft / 1 hPa
So Pressure at FL160 is 1013,25. Pressure difference is 10 hPa. So difference is 270ft.
16000ft - 270ft = 15730ft
After that you have to do temperature correction. ISA Temperature in FL160 would be -17°C. OAT is -27°C. Difference is 10°C.
Formula for temperature correction is 0,4% per 1°C: 0,4% * 10°C = 4% of 15730ft = 629ft which have to be substracted from 15730ft because it's colder than ISA.
So result must be 15101ft.
Hopefully. If not please correct me.
Oh now I see that your qquestion was answered in a previous thread? so why dou you ask again? :-)
first you have to do pressure correction with: 27ft / 1 hPa
So Pressure at FL160 is 1013,25. Pressure difference is 10 hPa. So difference is 270ft.
16000ft - 270ft = 15730ft
After that you have to do temperature correction. ISA Temperature in FL160 would be -17°C. OAT is -27°C. Difference is 10°C.
Formula for temperature correction is 0,4% per 1°C: 0,4% * 10°C = 4% of 15730ft = 629ft which have to be substracted from 15730ft because it's colder than ISA.
So result must be 15101ft.
Hopefully. If not please correct me.
Oh now I see that your qquestion was answered in a previous thread? so why dou you ask again? :-)
Last edited by Frolic; 12th Oct 2005 at 22:01.
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Number Cr.
Number cruncher I have just looke dat my school notes and seen the connection with 4 per 1000 - u must think im an idiot sorry- justbeen inundated with loads of formulaes and overlooked this one ...many thanks for your help
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yes I know but my result (15101) is quite near to the result in the feedback sheet.
So we were told that we have to do pressure correction first and afterwards the temp correction. JAR probably first temp corrects and pressure corrects afterwards. If you do it in this manner you'll get 15090.
But why do you aks this question again? It has been answered in a prevouis thread.
So we were told that we have to do pressure correction first and afterwards the temp correction. JAR probably first temp corrects and pressure corrects afterwards. If you do it in this manner you'll get 15090.
But why do you aks this question again? It has been answered in a prevouis thread.
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Gen Nav question, Heeeeelp?
Hi all, anyone figure this one out for me?
A route is being plotted on a Lamberts Chart using a grid aligned with the Greenwich meridian. The aircraft is at 49 S 100 E tracking 100 G. The convergence factor for the chart is 0.75, variation is 10 E, deviation is 4 E and drift is 7 right. The aircraft s true track is:
(a) 011
(b) 008
(c) 018
(d) 025
Ans is 25, but is my working out of it correct.
If the convergence factor is .75 and the chlong between datum and 100E is 100dgs, then the convergency between datum and 100E is 75dgs.
Diff between grid and true track is the convergency between the datum meridian and that point, ie 75dgs, therefore 100 -75 is 25?
So all the deviation, drift and variation is to confuse you?
Be nice to have this clarified, thanks for any comments
Powdermonkey
My exams are next week....... and this one I simply can't get.
Would be thankfull if someone could show me how to approach this one.
Cheers
A route is being plotted on a Lamberts Chart using a grid aligned with the Greenwich meridian. The aircraft is at 49 S 100 E tracking 100 G. The convergence factor for the chart is 0.75, variation is 10 E, deviation is 4 E and drift is 7 right. The aircraft s true track is:
(a) 011
(b) 008
(c) 018
(d) 025
Ans is 25, but is my working out of it correct.
If the convergence factor is .75 and the chlong between datum and 100E is 100dgs, then the convergency between datum and 100E is 75dgs.
Diff between grid and true track is the convergency between the datum meridian and that point, ie 75dgs, therefore 100 -75 is 25?
So all the deviation, drift and variation is to confuse you?
Be nice to have this clarified, thanks for any comments
Powdermonkey
My exams are next week....... and this one I simply can't get.
Would be thankfull if someone could show me how to approach this one.
Cheers
Jet Blast Rat
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Cadbury's Dairy Milk Very Tasty Creamy and Good
C D M V T C G
You have been asked to convert from grid to true heading. From the standard table above, remembered by mnemonics similar to the one I have given, the difference is (grid) convergence, which is 75°E.
Variation would then give you magnetic heading, deviation subsequently compass heading. But you have not been asked for either! Drift would tell you heading, but you've not been asked for that either. Read the question and answer it, don't try to go any further! Those three pieces of information are complete red herrings, they are of no use in answering the question. They give you such infoirmation o fool you in many exam questions, especially those concerning different bearings like this one.
Good luck!
Send Clowns
Gen Nav, BCFT
C D M V T C G
You have been asked to convert from grid to true heading. From the standard table above, remembered by mnemonics similar to the one I have given, the difference is (grid) convergence, which is 75°E.
Variation would then give you magnetic heading, deviation subsequently compass heading. But you have not been asked for either! Drift would tell you heading, but you've not been asked for that either. Read the question and answer it, don't try to go any further! Those three pieces of information are complete red herrings, they are of no use in answering the question. They give you such infoirmation o fool you in many exam questions, especially those concerning different bearings like this one.
Good luck!
Send Clowns
Gen Nav, BCFT
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First you must find the QNH correction for two purposes -
(1) Establish the Sea Level Pressure Height to ascertain the ISA Temperature there, and
(2) Establish the Indicated Altitude, upon which column of air the Temperature deviation affects in the Indicated to True Altitude conversion.
For a QNH of 1003, the Pressure Height at Sea Level is 281.1 feet. ISA Temperature at 281.1 feet is +14.45°C.
At Pressure Height 16000 feet, ISA temperature is -16.69°C, and for OAT = -27°C, Temperature is ISA-10.31°C.
At Pressure Height 16000 feet, for QNH = 1003, Indicated Altitude = 16000-281.1=15718.9 feet.
The Mean ISA Temperature of the column of air is (+14.45-16.69)/2 =-1.12°C + 273.15 =272.03°K.
The Mean Actual Temperature of the column of air is 272.03°K - 10.31 = 261.72°K.
True Altitude = 15718.9 X 261.72 / 272.03 = 15123.2 feet.
That's very close to the standard answer of 15090 feet.
The results arrived at here were achieved using actual almospheric values in place of the commonly used convenient approximations. The degree of complexity was not intended as a "line shoot", but to make three points -
(1) The practical day to day approximations used (30 ft/hPa, 2° per 1000 feet, etc.), whilst not exact, are good practical figures and may be used with confidence.
(2) For Examiners - Optional answers provided MUST be exact, with allowance for small deviation from the 'absolutely correct' answer. It can be extremely distressing to an examinee who has worked his/her calculation thoroughly, and finds that none of your answers 'fit'.
(3) For Students - It's essential in your learning process that you seek to understand WHY certain processes are involved, instead of the all-too-common approach of simply learning how to do it without any understanding of the factors involved. An understanding of WHY certain phenomena occur will last you for decades, whereas the simple rote learning of how to do it will be receding in your mind the day after the examination.
Regards,
Old Smokey
(1) Establish the Sea Level Pressure Height to ascertain the ISA Temperature there, and
(2) Establish the Indicated Altitude, upon which column of air the Temperature deviation affects in the Indicated to True Altitude conversion.
For a QNH of 1003, the Pressure Height at Sea Level is 281.1 feet. ISA Temperature at 281.1 feet is +14.45°C.
At Pressure Height 16000 feet, ISA temperature is -16.69°C, and for OAT = -27°C, Temperature is ISA-10.31°C.
At Pressure Height 16000 feet, for QNH = 1003, Indicated Altitude = 16000-281.1=15718.9 feet.
The Mean ISA Temperature of the column of air is (+14.45-16.69)/2 =-1.12°C + 273.15 =272.03°K.
The Mean Actual Temperature of the column of air is 272.03°K - 10.31 = 261.72°K.
True Altitude = 15718.9 X 261.72 / 272.03 = 15123.2 feet.
That's very close to the standard answer of 15090 feet.
The results arrived at here were achieved using actual almospheric values in place of the commonly used convenient approximations. The degree of complexity was not intended as a "line shoot", but to make three points -
(1) The practical day to day approximations used (30 ft/hPa, 2° per 1000 feet, etc.), whilst not exact, are good practical figures and may be used with confidence.
(2) For Examiners - Optional answers provided MUST be exact, with allowance for small deviation from the 'absolutely correct' answer. It can be extremely distressing to an examinee who has worked his/her calculation thoroughly, and finds that none of your answers 'fit'.
(3) For Students - It's essential in your learning process that you seek to understand WHY certain processes are involved, instead of the all-too-common approach of simply learning how to do it without any understanding of the factors involved. An understanding of WHY certain phenomena occur will last you for decades, whereas the simple rote learning of how to do it will be receding in your mind the day after the examination.
Regards,
Old Smokey
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Hi, the way I saw it was this.
First correct for barometric error, 10 x 27 = 270ft
So your actual alt is 15730ft.
Temp at 15730 should be in ISA -16.5 dgs C
Temp is -27 at 15730ft so at that level its ISA -10.5
.4% for every dg below isa leaves at correction of 4.2%
15730 - (4.2% of 15730) = 15070ft
Is this correct? I am not sure why you guys were working out the ISA temp deviation at FL160, it's not your actual altitude. Is this wrong on my part? Should you first do the temp correction then barometric?
First correct for barometric error, 10 x 27 = 270ft
So your actual alt is 15730ft.
Temp at 15730 should be in ISA -16.5 dgs C
Temp is -27 at 15730ft so at that level its ISA -10.5
.4% for every dg below isa leaves at correction of 4.2%
15730 - (4.2% of 15730) = 15070ft
Is this correct? I am not sure why you guys were working out the ISA temp deviation at FL160, it's not your actual altitude. Is this wrong on my part? Should you first do the temp correction then barometric?
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powdermonkey, ISA Temperature (and deviation) MUST be evaluated for the PRESSURE HEIGHT. Having found the Temperature deviation from ISA, the correction is then applied to the ALTITUDE.
Regards,
Old Smokey
Regards,
Old Smokey
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Thanks Old Smokey
I won't forget that bit of info, I am surprised that with the amount of such questions I have done, that this particular problem has only come up now!? Still better to find this out a few days before my exams than a few days after................so much still to learn............and I'm not getting any younger...
Still, it's more fun than doing my old job
Cheers
I won't forget that bit of info, I am surprised that with the amount of such questions I have done, that this particular problem has only come up now!? Still better to find this out a few days before my exams than a few days after................so much still to learn............and I'm not getting any younger...
Still, it's more fun than doing my old job
Cheers
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As number cruncher stated, what you will need to remember for the exam are these two formulas:
1. First two digits of flight level x Isa Deviation x 4
Then to solve the last bit of this particular question as the QNH is not 1013.25:
2. QNH deviation *27 - (Add or subtract depending on whether QNH is Higher or Lower)
These will get you closest to what the JAA deem the correct answer
1. First two digits of flight level x Isa Deviation x 4
Then to solve the last bit of this particular question as the QNH is not 1013.25:
2. QNH deviation *27 - (Add or subtract depending on whether QNH is Higher or Lower)
These will get you closest to what the JAA deem the correct answer
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geez glad i asked this question -- see the response!!
agree with powders point as Met manual does not specify press alt calculation - ie crp5 usage,,, just reg 4per 1000ft rule
agree with powders point as Met manual does not specify press alt calculation - ie crp5 usage,,, just reg 4per 1000ft rule
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RNAV question CDI from/to
A Course Dev Indicator for a VOR selected to 90 deg
From /To indicates ''To''
CDI Needle is deflected halfway to right
What is radial of Aircraft?
a 095
b 275
c085
d 265
I made it 265 as if needle is deflected to right the ac has to follow needle until centred- am i right? so nearest deg is 085deg
radial of 085 is 265 answer (d)
In my notes the feedback ans is 275 ... how come?
From /To indicates ''To''
CDI Needle is deflected halfway to right
What is radial of Aircraft?
a 095
b 275
c085
d 265
I made it 265 as if needle is deflected to right the ac has to follow needle until centred- am i right? so nearest deg is 085deg
radial of 085 is 265 answer (d)
In my notes the feedback ans is 275 ... how come?
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The VOR OBS on the instrument is set to 090 degrees and has a "To" flag displayed.
All Radials are "FROM" the VOR.
This means that although the instrument in the aeroplane is set to 090 the actual radial set on the instrument is the 270R. You can see this by twisting the OBS to show 270 at the top of the instrument. You will then have a "FROM" flag.
A halfway deflection on the instrument is 5 degrees (2 degrees per dot on the instrument face for VORs).
So as the needle is deflected right you are flying to the left of track. If you are to the left of track that will put you on a radial greater than 270R. 270R plus 5 degree deflection = you are flying on the 275 Radial.
All Radials are "FROM" the VOR.
This means that although the instrument in the aeroplane is set to 090 the actual radial set on the instrument is the 270R. You can see this by twisting the OBS to show 270 at the top of the instrument. You will then have a "FROM" flag.
A halfway deflection on the instrument is 5 degrees (2 degrees per dot on the instrument face for VORs).
So as the needle is deflected right you are flying to the left of track. If you are to the left of track that will put you on a radial greater than 270R. 270R plus 5 degree deflection = you are flying on the 275 Radial.
Jet Blast Rat
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You don't follow needles on a CDI to intercept the inbound track - you go further than the indicated deviation!
The indication is saying that you have to track right of 090° to intercept the 090°M inbound track, and that you are 5 degrees off. Therefore if you track 5° right of 090° you are directly inbound, so you are on a 095°M inbound track, or QDM. As CZ says radials are outbound magnetic bearings (QDRs) so you are on the reciprocal, 275°.
The indication is saying that you have to track right of 090° to intercept the 090°M inbound track, and that you are 5 degrees off. Therefore if you track 5° right of 090° you are directly inbound, so you are on a 095°M inbound track, or QDM. As CZ says radials are outbound magnetic bearings (QDRs) so you are on the reciprocal, 275°.
Jet Blast Rat
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Pressure altitude is required for perfect accuracy. However the difference it makes is really rather small, and you will find it is not given in some questions in some schools' question banks and feedback. Not a major issue - look how close together the PA marks are on the CRP-5, then remember that the QNH is rarely going to make more than 500' difference.
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Ground School Exam Questions & Question Banks
Hi All,
I'm thinking of paying the dosh for access to the Bristol question bank. First thing, I've noticed they have increased he price quite drastically. Secondly I've been told they have around;
POF - 444 Q's
AGK - 764 Q's
Perf - 361 Q's
Is this this correct?
Cheers.
RB
I'm thinking of paying the dosh for access to the Bristol question bank. First thing, I've noticed they have increased he price quite drastically. Secondly I've been told they have around;
POF - 444 Q's
AGK - 764 Q's
Perf - 361 Q's
Is this this correct?
Cheers.
RB
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Do not know about the amount of questions in the bank, but the cost is still one of the best around
http://www.pprune.org/forums/showthr...=question+bank
See
YYZ
http://www.pprune.org/forums/showthr...=question+bank
See
YYZ