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Supercharged PPRuNer
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(b) increases, remains constant.
Descent at constant mach # means increasing IAS, thus decreasing CL and increasing descent angle. Switching to constant IAS means constant CL, and constant descent angle.
Do I win £5?
Descent at constant mach # means increasing IAS, thus decreasing CL and increasing descent angle. Switching to constant IAS means constant CL, and constant descent angle.
Do I win £5?
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Yep, I'd go for B.
Think this question is attempting to confuse you with angle of descent and angle of attack.
Descending with a constant Mach no. will see an increase in EAS, which will result in the need for a lower angle of attack. If you lower the angle of attack in a descent the nose will go down, and if the a/c nose goes down then it goes without saying that the angle of descent will increase.
Think this question is attempting to confuse you with angle of descent and angle of attack.
Descending with a constant Mach no. will see an increase in EAS, which will result in the need for a lower angle of attack. If you lower the angle of attack in a descent the nose will go down, and if the a/c nose goes down then it goes without saying that the angle of descent will increase.
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I concur - B is the answer to go for. A quick(!) explanation off the top of my head:
Mach Number is the ratio of true airspeed to the local speed of sound. Local speed of sound is solely dependent on temperature, which decreases with increasing altitude. So as you descend, the air temperature is increasing and therefore the local speed of sound is increasing. To maintain the same mach no the indicated (and therefore true) airspeed will have to increase, which means you lower the nose and increase the descent angle.
As you descend further you will reach an altitude where the mach number you are descending are coincides with the indicated airspeed you want to descend at. If you keep on lowering the nose after this point you will exceed this airspeed and so the descent angle is held constant.
Hope that helps!
Al
Mach Number is the ratio of true airspeed to the local speed of sound. Local speed of sound is solely dependent on temperature, which decreases with increasing altitude. So as you descend, the air temperature is increasing and therefore the local speed of sound is increasing. To maintain the same mach no the indicated (and therefore true) airspeed will have to increase, which means you lower the nose and increase the descent angle.
As you descend further you will reach an altitude where the mach number you are descending are coincides with the indicated airspeed you want to descend at. If you keep on lowering the nose after this point you will exceed this airspeed and so the descent angle is held constant.
Hope that helps!
Al
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Please don't grill me as I've tried to fid the details in GETMET and other sources but what does the AUTO and NDV mean in this METAR as I haven't seen this before.
EGPU 231450Z AUTO 20003KT 9999NDV // BKN022 SCT044 10/04 Q1004
Thanks for your help
The Reverand
EGPU 231450Z AUTO 20003KT 9999NDV // BKN022 SCT044 10/04 Q1004
Thanks for your help
The Reverand
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AUTO (surprise surprise), this METAR was produced by a machine, rather than by a person.
NDV means "No Directional Variations" available. Ordinarily, in conditions of restricted visibility, an observer may indicate the direction in which the visibility is either best or worst. In this case, no such distinction is available. 9999 in all directions.
2D
NDV means "No Directional Variations" available. Ordinarily, in conditions of restricted visibility, an observer may indicate the direction in which the visibility is either best or worst. In this case, no such distinction is available. 9999 in all directions.
2D
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At CAS 250kts, FL390, JSA +8*C what is the TAS?
A)439
B)482
C)458
D)464
The ans is given as 464kts, but I keep getting 482... the only way to get to 464 is to use ISA values for the temp, am i right in thinking the temp in this question is
-55? or have i messed up because its the JSA???
Confussed???!!!
A)439
B)482
C)458
D)464
The ans is given as 464kts, but I keep getting 482... the only way to get to 464 is to use ISA values for the temp, am i right in thinking the temp in this question is
-55? or have i messed up because its the JSA???
Confussed???!!!
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I suspect that you are not applying the correction for compressibility.
JSA +8 @ FL390 is -55ºC, as you say. That gives 482 knots before correction for compressibility (with a CRP-5 - you may find that an ARC2 gives a slightly different answer).
With compressibility corrected,the answer comes to about 463 or 464.
If your FTO has not shown you how to correct for compressibility, read the little booklet that comes with the flight computer.
JSA +8 @ FL390 is -55ºC, as you say. That gives 482 knots before correction for compressibility (with a CRP-5 - you may find that an ARC2 gives a slightly different answer).
With compressibility corrected,the answer comes to about 463 or 464.
If your FTO has not shown you how to correct for compressibility, read the little booklet that comes with the flight computer.
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inst question
the interception of a ILS beam by the auto pilot takes place :
A)according to a interception verces range and angle
B)at a constant heading
The white arc of the EGT gauge is:
A)special op range
b)normal op range
A)according to a interception verces range and angle
B)at a constant heading
The white arc of the EGT gauge is:
A)special op range
b)normal op range
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Hi there,
1. a
2. a
Been a while since I studied for the Instruments exam, but I do remember these questions and I'm fairly confident in my answers.
Please correct me if I'm wrong
1. a
2. a
Been a while since I studied for the Instruments exam, but I do remember these questions and I'm fairly confident in my answers.
Please correct me if I'm wrong
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The first one is (a).
Think of the special operating range of an ASI. That'd be the Flap operating range... what colour is it? White. Apply this concept to the EGT guage. (That is how I remember this one anyway).
The second is also (a).
Think of the special operating range of an ASI. That'd be the Flap operating range... what colour is it? White. Apply this concept to the EGT guage. (That is how I remember this one anyway).
The second is also (a).
Last edited by Charlie Zulu; 28th Jun 2005 at 15:43.
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a and a...
when you intercept the ILS, your plane is using a sinusoidal signal and calculate the angle to intercept the ILS .angle are different for any distance.So it can not be at straight angle.
when you intercept the ILS, your plane is using a sinusoidal signal and calculate the angle to intercept the ILS .angle are different for any distance.So it can not be at straight angle.
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anyone answer this for me ? and why?
for a jet cruising at 1.32 vimd a 5% decrease in weight would give a change of fuel flow by approx:
10% decrease
5% decrease
5% increase
2.5% decrease
cheers..
for a jet cruising at 1.32 vimd a 5% decrease in weight would give a change of fuel flow by approx:
10% decrease
5% decrease
5% increase
2.5% decrease
cheers..
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1.32 vmd
Fuel flow at 1.32 vmd will decrease at the square root of the change in weight or (for small changes half the change.)
hense 2.5%
At VMD the change would be equal
At VMP the change would be squared(doubled for small change)
Fullrich
hense 2.5%
At VMD the change would be equal
At VMP the change would be squared(doubled for small change)
Fullrich
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Thats very interesting Fullrich and makes sense when you consider the shape of the drag curve
I have read much debate on similar questions and your reply has clarity
Have you a reference for it?
I have read much debate on similar questions and your reply has clarity
Have you a reference for it?
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hey guys and gals ...
I faild instruments becasue of this question. Does anyone know how 2 answer it, because my father and i and a few other guys have no clue!!
here it goes ...
An aircraft flies a mesured course of 5NM between two pylons at 7000ft palt ., temp 15deg celcisus in 2min45sec. flying the revers in 2min19sec
if the asi was 100kts the asi error was
(A) 1kts under read
(B) 3kts over read
(C) 3kts under read
thanx hope sumone can help
sean
I faild instruments becasue of this question. Does anyone know how 2 answer it, because my father and i and a few other guys have no clue!!
here it goes ...
An aircraft flies a mesured course of 5NM between two pylons at 7000ft palt ., temp 15deg celcisus in 2min45sec. flying the revers in 2min19sec
if the asi was 100kts the asi error was
(A) 1kts under read
(B) 3kts over read
(C) 3kts under read
thanx hope sumone can help
sean
Last edited by zssp; 13th Aug 2005 at 20:02.
Warning Toxic!
Disgusted of Tunbridge
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Are you certain you have made no typographical errors in the numbers?
I think it is begging for this:
Total track miles......... 10 miles.
Total time...................2.45 + 2.15=5.00
Therefore TAS 120kts ->
A circular slide rule should make this easy- it's not to hand, so I have had to resort to complicated corrections below.
From "Mathematically increase your indicated airspeed (IAS) by 2% per thousand feet of altitude to obtain the true airspeed (TAS)." and to allow for temperature error, "when changing IAS to TAS you need to add one knot to your Indicated airspeed (IAS) for each five or six degrees the temp is above standard day, at altitudes from 5,000 to 15,000 ft"(http://www.csgnetwork.com/tasinfocalc.html) and (http://www.sonic.net/~pjkelly/tswinds.html)
So, TAS is higher than IAS by 14% for Pressure Altitude correction, or Pressure corrected IAS=106 kts(106+14%=120), then subtract 3 kts for temperature error (because ISA at 7000' is +1 deg C, ie we are flying at ISA+14deg C), ie IAS should be 103kts, but it is actually 100kts, so it is underreading by 3.
Now PLEASE can we have a look at your grammar?
I think it is begging for this:
Total track miles......... 10 miles.
Total time...................2.45 + 2.15=5.00
Therefore TAS 120kts ->
A circular slide rule should make this easy- it's not to hand, so I have had to resort to complicated corrections below.
From "Mathematically increase your indicated airspeed (IAS) by 2% per thousand feet of altitude to obtain the true airspeed (TAS)." and to allow for temperature error, "when changing IAS to TAS you need to add one knot to your Indicated airspeed (IAS) for each five or six degrees the temp is above standard day, at altitudes from 5,000 to 15,000 ft"(http://www.csgnetwork.com/tasinfocalc.html) and (http://www.sonic.net/~pjkelly/tswinds.html)
So, TAS is higher than IAS by 14% for Pressure Altitude correction, or Pressure corrected IAS=106 kts(106+14%=120), then subtract 3 kts for temperature error (because ISA at 7000' is +1 deg C, ie we are flying at ISA+14deg C), ie IAS should be 103kts, but it is actually 100kts, so it is underreading by 3.
Now PLEASE can we have a look at your grammar?
Last edited by Rainboe; 13th Aug 2005 at 19:56.
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thanx .. thats how i was asked the question in the paper and ive found the same one in 2 question books thanx for the help .. i looked all over for a way to work it out and you way seems to make more sense ... i do appologise i shuld type slower and read over what ive writen
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(Head in hands....crying)
It's actually 34 years since I did this sort of calculation, so any confirmation from anybody would be very welcome (but do start sentences with capitals and don't forget punctuation!).
It's actually 34 years since I did this sort of calculation, so any confirmation from anybody would be very welcome (but do start sentences with capitals and don't forget punctuation!).
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Well it was really good i must say. I really do think thats an ATPL question. I know my problem is i just try get things done in to much of a rush. But i will try more in future but thanx so much for helping me out.
hey a guy on avcom wrote this
i would choose B - 3 knots over-read. A calibrated airspeed of 100Kts at 15 C at a Palt of 7000 feet equates to 113 Kts TAS. As the different times taken to fly the route indicates a wind correction of 10 Kts needs to be applied (129 Kts one way and 109 kts returning) Hence the TAS -Wind correction equals 103 knots - hence 3 knot over-read
b is the answer in the book
hey a guy on avcom wrote this
i would choose B - 3 knots over-read. A calibrated airspeed of 100Kts at 15 C at a Palt of 7000 feet equates to 113 Kts TAS. As the different times taken to fly the route indicates a wind correction of 10 Kts needs to be applied (129 Kts one way and 109 kts returning) Hence the TAS -Wind correction equals 103 knots - hence 3 knot over-read
b is the answer in the book