Ground School Exam Questions & Question Banks
 

an aircraft must clear an obstacle at 2000ft elevation by 1000ft. Regional QNH is correctly set at 998mb, the OAT at 3000ft indicated is -10C. At what indicated altitude will the 1000ft clearance be achieved?

Explaination please

This is a straightforward indicated to true altitude. Use your CRP-5 by preference. Otherwise, 4 feet per 1000 feet amsl per degree off ISA. 3000 amsl, 998 HPa is roughly 3500' (3400', but that is too precise) Pressure Altitude so ISA is 8 degrees. The temperature is therefore 18 degrees cold, so the correction is 4 x 18 x 3 = 216 feet. The weather is cold so altimeter is over-reading, so you are clear at 3216 feet indicated.

Send Clowns
Gen Nav
BCFT

P of F question
 

By what percentage does VA (EAS) alter when the aeroplanes weight decreases by 19%?

a)4.36% lower
b)No change
c)19% lower
d)10% lower

Answer is d. how is this worked out?

Thanks in advance

Foz

If old weight were 1000 then new weight would be 1190.

SQRT(1190/1000) = 1.091 (or approx 10%)

Think about the classic PPL exam question of the stall speed at 60 deg bank in level flight. The force would be 2g or, in other words the weight doubles. SQRT(2) = 1.41...stall speed increases by 41%.

So it would actually be sqrt 810/1000 as the weight is being decreased. Comes out at 0.9 (90%) which is 10% lower! Perfect.

Cheers

Foz

It is a quirk of the mathematics that for small changes of weight - less than about 20% - the percentage change of speed is about half the percentage change in weight.

Dick W

VA is the stalling speed at the limiting load factor. So this question concerns the way in which stalling speed changes with changes in weight.

The general solution for finding the change of stalling speed for any given change in weight is,

Vs at new wt = Vs at old wt x sguare root (new wt/old wt)

As Dick said, for small weight changes the % change in Vs is about half the % change in weight.

Even for a 50% increase in weight this approximation gives 25% increase whereas the actual figure is 22.47%, so it is still reasonably close.

Yeah, that's what I meant :O

Performance Question
 

Hy

can anyone help me with this one?

For a class B a/c landing on wet grass the minimum acceptable LDA is..... times the LD?

a. 1.67
b. 1.77
c. 1.87
d. 1.97

official answer: d

my answer c.

Class B a/c means 70% -> 1.428
Wet Rwy: -> 1.15
grass: -> 1.15

This should give a Factor of 1.88!
Obviously I did something wrong, and therefore need some assistance!
Thanks a lot.

I think it should be 1.89. But 1.87 is the nearest best answer. So I would have chosen that in an exam.

Would agree with High Wing Drifter!!

LDA >or= LDR * Wet factor * Surface factor * 1.43
LDA >or= LDR * 1.15 * 1.15 * 1.43
LDA >or= LDR * 1.89

The selection of 'd' may be based on the following somewhat odd logic.

The calculated minimum is LDR * 1.15 * 1.15 * 1.43 = LDR*1.891175.

Of the four choices presented, the only one which meets the required minimum is (d). So if you had to choose one of those four multipliers, you'd have to go with 1.97, because 1.87 is below the required minimum.

I think it's a stupid logic trap if they intended that, because the question is poorly worded also. But that might be thought process involved....

Spot on Mad. Serves me right for jumping to a rash conclusion :\

ATPL met question
 

Hi Guys. im studying for the ATPL by distance learning (just started). and im tearing my hair out on a particular question. Its been a long day and maybe im being a doofus, but maybe you guys can point me in the right direction. I dont expect you to give the answer, id only be cheating myself!

a 10,000 foot mountain has a 5,000 foot cloudbase to windward and a 7,000 cloudbase to the lee. if the windward temp is +20 degrees C what would be the likely temp to the lee?

now my working was, assuming temp decrease @ DALR to 5,000, then decrease @ SALR to 10,000, then increase @ salr to 7000, then increase @ DALR back to surface, you come up with+23 degrees C !
the options for answers are
a-22.4
b-16.8
c-20.4
d-25.6 !!!!!
where the smeg am I going wrong ?!
of the possible answers 22.4 is the closest, do they add a little for the temp drop over the top of the mountain? however much that may be? or have I got some basic thing wrong? is there anything in the word 'likely' temp? that doesnt sound too specific !!!

am really enjoying my study by the way!!!

Hi,

Just a guess: If you use 1.8 for SALR and 3 for DALR, it gives you answer A. Bristol ground school uses these values. Hope this helps.

edit: sorry, I didn't read your post carefully, you didn't wan't the answer :O

thanks guys. the company im with says 3 degrees C for DALR and 1.5 degrees SALR. also, Pressman.....how??? I did that and got 23 degrees.
temp at 5000 feet windward 5 degrees, decreasing to minus 2.5 at 10,000. then back to plus 2 at 7000 feet on the lee and increasing at DALR (3degrees per 1000 feet) back to surface. i just cant get the point something or other, can you show how ? many thanks once again.

Exam technique - You have to select the most correct answer, or the least wrong one! Otherwise it's possible to get the right answer without doing the working.

Phil

Your working should look something like this.

20-(5x3)=5-(5x1.8)=-4+(3x1.8)=1.4+(7x3)=22.4

It is important to remember that the SALR is not constant as we gain altitude. Latent heat is released as condensation occurs and the SALR value actually increases and eventually could equal the DALR

We can debate this till the cows come home but at the end of the day the CAA say that the SALR is 1.8 degrees per thousand feet.

Your thinking was perfectly correct. 1.8 is the number to remember.

fohn wind
 

This is easier!

Warming temperature =

1.5 times the difference in cloud bases ( in thousands of feet),

7000 - 5000 = 2 x 1.5 = 3

Add to the temperature on the windward side, 3 + 20 = 23.

You'll never get closer than that without wasting valuable exam time!

All donations gratefully received.........

Time!
 

This type of question for JAA ATPL is a waste of time, so trying to do it quicker is a waste of time too. If JAA ever ask this type of question, I'll eat my hat!.
All you need to know about Foehn for the JAA is this.
Warmer on ther lee side, higher cloud on the lee side, possible turbulence on lee side (from possible mountain waves) and you need to know the other names for the Foehn, e.g Santa Ana and Chinook.
Now for some lovely spuriouis info. The Foehn translated actually means "Hair dryer" in some European languages, so remember it as a warm dry descending wind, or a warm katabatic.

One more thing, if your School is using 1.5°C per 1000ft for the SALR in their ATPL notes they need shooting and they need to be reminded of the JAA METEOROLOGY OBJECTIVES.

Good luck anyway, met can be a nightmare unless your taught right, wait till you get to winds!!!

Have fun