The theory is that you can determine whether you are flying from/to a higher or lower pressure or temperature based on the direction of crosswind. This is Buys Ballot's law. In the Northern Hemisphere, if you stand with your back to the wind, the low pressure system will be on your left. The reverse is true in the Southern Hemisphere. Therefore if you have a crosswind from the left (ie. starboard drift), you are flying towards a lower pressure. Then remember the phrase: "high to low, watch out below!", which will remind you that true altitude decreases as you fly towards a lower temperature or pressure.

Wow ok I get it!! Just clicked. THANK YOU!

need help with a question
 

Friends, i am stuck with one of a question.can some plz shed some light in solving this question.thanks for the help
A)18° B)22° C)20° D)24°
ANS IS B.

Tried it with:

cloud base = 400x (T-Td)
SALR= 1.8 degrees/ 1000 ft
DALR= 3 degrees/ 1000 ft

cools with DALR up to cloud base on windward side, then cools with SALR until the summit, then warms with SALR as it descends on the leeward side until cloud base, then warms with DALR again.

But calculating this gives 20 degrees:rolleyes:

Mass and Balance Question
 

Hi everyone A mass and balance question I need help with!

At a given mass the CG position is at 15% MAC. If the leading edge of MAC is at a position 625.6 inches aft of the datum and the MAC is given as 134.5 inches determine the position of the CG in relation to the datum.

Withn the answer it states -
The MAC is 134.5" long. The CG is 15% of this distance back from the leading edge. 15% of 134.5" is 20.17".
The leading edge of the MAC is 625.6" aft of the datum, the CG is 625.6 + 20.17 = 645.77" aft.

WHERE DO THEY GET 20.17 from!!

Or am I just being DUMB!

Length of MAC = 134.5, so:

1% mac = 134.5/100 =1.345
15% mac = 1.345 * 15 = 20.175

You need to understand the meanings of the various terms.

The MAC is the Mean Aerodynamic Chord Length.

The statement that means the distance from the MAC leading edge to the MAC trailing edge is 134.5 inches.

15% of 134.5 inches is 20.175 inches.

So if the CG is at 15% MAC it is 20.175 inches aft of the MAC leading edge.

If we now add the MAC leading edge poistion of 625.6 inches aft of datum we get a CG position of 20.175 + 625.6 = 645.775 inches aft of datum.

Thank you............
 

..........Mr. Williams for taking the time to so clearly answer the original question.

Pitot questions on recent ATPL exams
 

Hello all.

I'm looking for recent ATPL questions regarding Pitot blockages. There seems to be some discrepancies between our schools studyguides and real life. As I am a teacher in the subject, it would be interesting to know which stance the current QB has.

Our studyguide maintains that a blockage of the ram air opening of a pitot tube will cause the ASI to freeze at current speed whereas in real life, such a blocking would cause the ASI to read 0 because of pressure escaping through the waterdrain.

Any thoughts?

Try it.

Go to an aeroplane with a friend - blow GENTLY into the pitot, then hold your finger firmly over the end. Then have your friend go around and see what is happening in the cockpit.

I know what I've got every time I've done that -but try it yourself and you'll never forget the answer.

G

......and I have been for twenty years.

Suggest you teach correctly........

The altimeter will freeze if a blockage occurs but that would be in the static system. The VSI would read zero. If the pitot gets blocked, the ASI behaves like an altimeter and will increase its readings as you climb. If the static gets blocked, the ASI error will reverse, i.e. it will under-read as you climb.

At least that's what the JAA questions expect - real life has nothing to do with it! :)

The water drain is usually operated with a spring (every 200 hours on some aircraft), so that implies some sort of seal. It would have to be sealed in normal circumstances (like the alternate static) otherwise the readings would never be correct.

Two times in my life I've had a blocked pitot.

One it did pretty much as the JAA exams expect - froze at the original value (I never went above about 2000ft that trip, so the altimeter thang wasn't really there).

The other did something I'd never have predicted - it started reading almost in direct proportion to the RPM gauge. During take-off of-course, that actually seems about right - after five minutes I realised that something odd was happening.

Fortunately both totally VFR trips, so all flown on attitude back to an uneventful landing.

The "blow & block" test I've learned now to do as part of my pre-flight whenever I'm doing the first flight of a new-build aeroplane ! It's also why I know that this cobblers about water drains is just that - if it starts going down, there's a system leak and it needs sorting before flight.

It's a problem with "teachers" who have no real knowledge or experience - but aviation suffers from that a lot (as do many schools of-course, so they're in good company).

G

Blow and block technique works for me - but you do need two people and be aware when the last flight happened; pitot heat ruins your chances in the pub later. :(

How do you calculate the lowest useable flight level?

a. Lowest QNH and lowest negative temperature below ISA
b. Lowest QNH and highest negative temperature below ISA
c. Highest QNH and highest temperature above ISA
d. Highest QNH and lowest temperature


Please Help.

What temperature and pressure conditions would be safest to ensure that your flight level clears all the obstacles by the greatest margin?

a. Cold temp/low pressure
b. Warm temp/high pressure
c. Temp less than or equal to ISA and a QNH less than 1013
d. Temp more than or equal to ISA and a QNH greater than 1013

The worst place to be is in a cold low.

Mass and Balance Questions - Traffic Load
 

I wonder if there are any brain boxes out there who can shed some light on this question? I have (as far as I am aware) followed the correct proceedure of how to calculate a Traffic Load.

MTOM: 170 000kg
ZFM: 112 500kg
MLM: 148 500kg
DOM: 80 400kg
TAXI FUEL: 800KG
BLOCK FUEL: 40 000KG
TRIP FUEL: 29 000KG.

Question asks for the traffic load which can be carried.
Here are the choices:
A) 32 100kg
B) 32 900kg
C) 18 900kg
D) 40 400kg

My calculations:

(*1) (I have calculated MTOM fuel to be block fuel (40 000kg) - taxi fuel (800kg) = (32 000kg) )

(*2) (I have calculated MLM fuel to be block fuel (40 000kg) - taxi fuel (800kg) - trip fuel (29 000kg) = 10 200kg

MTOM / ZFM / MLM
LIMIT 170 000kg / 112 500kg / 148 500kg
DOM 80 400kg / 80 400kg / 80 400kg
FUEL 39 200kg (*1) / ------- / 10 200kg (*2)

=TRAFFIC LOAD 50 400kg / 31 600kg / 59 900kg

THE CORRECT ANSWER SHOULD BE A but I don't get anywher near that!:confused:

Can anyone tell me where i have gone wrong? I thought that I had these mass and balance questions sorted :{

I'm currently studying with Bristol, but the question was taken from 'The Daily ATPL' website.

I would be so grateful if anyone could help me :)

The most limiting factor for this flight is:

MZFM - DOM = TL

112500 - 80400 = 32100 kg

Not sure how you got 31 600kg out of that calculation :ok:

Your problems started with your very first statement:

Re-check your maths - you're already almost 8 tonnes adrift...

ChriSat - thanks for your reply! I have no idea how I arrived at that either. I blame the calculator! I now see that I was working it out correctly, but I must have put in the wrong values or copied down the answer incorrectly.

pilotmike - thank you for your reply also. I see what you're saying, I meant to say 39 200kgs :oh:

PoF theory questions
 

Hi,

please help me to solve the following problems(providing an explanation):

1.If an airplane glides at an angle of attack of 10°, how much altitude will it lose in 1 mile?
A:240 feet.
B: 480 feet.
C: 960 feet.

2.How much altitude will this airplane lose in 3 miles of gliding at an angle of attack of 8°?
A:440 feet.
B: 880 feet.
C: 1,320 feet.

The figure to use: 3_2 | Flickr - Photo Sharing!

Thanks!

Mxms

The questions are based n the equation:

Glide Range = Height x L/D ratio

Rearranging this gives

Height = Glide Range / L/D ratio.

The diagram shows that

L/D ratio at 10 degrees angle of attack = approximately 11.2

L/D ratio at 8 degress angle of attack = approximately 12.2

If we assume that the 1 mile and 3 mile in the question are nautical miles and assuming 1 nm = 1680 feet we get

For 10 degrees height = 6080 / 11.2 = 542 ft

For 8 degrees height = 6080 x 3 / 12.2 = 1495 feet

These answers are not very close to the options given

But if we assume that the 1 mile and 3 mile in the question are statute miles (5280 feet) we get

For 10 degrees height = 5280 / 11.2 = 471 ft

For 8 degrees height = 5280 x 3 / 12.2 = 1298 feet

These are closer to the options of 480 feet and 1320 feet.

These are very unusual questions and are clearly not from the JAR ATPL system. Where did they come from?

Hello everybody

They are from the Peruvian CPL question bank. I asked Max for help with the question because I am here doing a license conversion I don't have any books with me.

Great answer, thanks a lot! :ok:

Cheers

Ola peruano

While you are here could you pse say how QNH is calculated at high altitude airfields like Cuzco and La Paz

Dick

ola gringo :)

I guess it's calculated just like at any airfield in the world, that is adjusted to sea level.

S

Gracias S

I had an idea that for high altitude fields a different datum was used.

No soy gringo, soy ingles - as I have had to say many times in Ecuador

Cheers, Dick

Don't mention it :)

I know you are ingles but I noticed that here they don't make the difference... it was meant as a joke, sorry about that.

Cheers
S

Altimetry Help
 

Good afternoon,

I think this is what it feels like to have mental block! I have just got 99 out of 100 altimetry questions correct, but I just can't work this one out for some reason. Maybe I'm staring too hard trying to find the right answer that I'm missing the obvious solution, but it's something I'd like to get clear in my head either way. You may ask why I'm so bothered when it's only one question out of 100 - well, because it's 'sods law' that it'll come up in the exam! This same question has been answered further up the thread, but it doesn't help solve the confusion in my case. Here goes:

"You plan a flight over a mountain range at a true altitude of 15000ft/AMSL. The air is on average 15C colder than than ISA, the pressure at sea level is 1003 hPa. What approximate altitude should the altimeter read (pressure setting 1013 hPa)?"

Now I'm quite good at remembering and re-arranging mathematical formulas as opposed to rules (just a personal preference), and so I use:

True Altitude = Altitude on QNH + [(ISA Deviation x 4) x (Pressure Altitude/1000)] and I've always got the answer to similar style questions spot on!

With the question above though, I can't get my head around which one of the missing components I'm actually looking for? We're given the True Altitude as 15,000ft, and I know that the ISA Deviation x 4 = -60, and so that means I either need to find the Altitude on the QNH or the pressure altitude, but I can't work out which.

Is anyone able to help explain using my formula? The answer by the way is supposedly 16,230', although I'm aware with these that the answers are quite often 50' out and you just have to pick the closest one.

Many thanks in advance.

Rearranging your equation

True Altitude = Altitude on QNH + [(ISA Deviation x 4) x (Pressure Altitude/1000)]

Gives

Alt on QNH = True Altitude - [(ISA Dev x 4) x (Pressure Alt/1000)]

Inserting the data provided gives

Alt on QNH = 15000 – ((-15 x 4 x 15) = 15900

But we are going to use 1013 which is 10 HPA higher than QNH.

Using 27 ft / Hpa this means that the 1013 level is 270 feet below msl

So adding 270 to 15900 gives an indicated altitude of 16170 feet.

At that altitude the estimated lapse rate of of 27 feet per Hpa is a bit too low. It looks as if the examiner has used 33 ft per HPa (30 HPa per 1000 ft) which is probably more realistic.

That's brilliant Keith, I think I've pretty much got it now!

Am I right in thinking that you're effectively working out what the Altitude on the QNH would be (1003 set) and so therefore Pressure Altitude is the same as True Altitude in the initial calculation? And then you're taking in to account the variation between sea level QNH and 1013?

Your equation isn't quite correct.

It should be

TA = IA + (4 x IA/1000 x ISA Dev) + (27 x (QNH – Subscale))

Where
TA = true altitude,
IA = indicated altitude,
Subscale = altimeter subscale setting,
ISA Dev = ISA temperature deviation

Strictly speaking the term "Pressure Altitude" means the altimeter reading when 1013 is set on the subscale. So if we were to use Pressure Altitude instead of indicated altitude in the above equation we would be applying the temperature error from the 1013 level upwards.

My use of TA in ,y previous post was an error. Using the above equation we can solve the problem as follows.

TA = IA + (4 x IA/1000 x ISA Dev) + (27 x (QNH – Subscale))


ISA Deviation = -15
True Altitude = 15000 ft
QNH = 1003

TA – (27 x (QNH-Subscale setting) = IA + ( (4 x IA x ISA Dev)/1000 )

Inserting the data provided in the question gives

15000 – ( 27 x (1003 – 1013) = IA + (4 x IA x (–15) ) / 1000 )

15000 + 270 = IA – 0.06 IA

15270 = 0.94 IA

15270 / 0.94 = IA

16244 = IA

So Indicated Altitude = 16244 feet

This is not quite the correct answer of 16230, but it is pretty close to it.

AC Electrics
 

Hey guys, I could use some input on this one.

If an alternator is run at below normal frequency, then:
A) Electric motors will stop.
B) Inductive devices will overheat.
C) Lights will become dim.
D) Lights will become brighter.

The correct(?) answer is B. I agree that if the frequency is below normal, that would mean that the inductive reactance is lower and thus the current will be higher, but conversely, a lower frequency would also mean that capacative reactance would be higher, thus giving a lower current, which should balance the lower inductive reactance, as there is no mention of one reactance being more dominant than the other, or any one kind of reactance at all for that matter, other than what might possibly be deducted when the answer alternatives are examined. Am I to assume that the other alternatives, being placed in a purely capacative reactant circuit couldn't occur in the case of decreased frequency?

-Anders

This is a case (just one of many) where the examiners have made some simplfiying assumptons, but have not stated what these assumptions were.

All they really want you to do is to note that reducing frequency reduces inductive reactance so for a given applied voltage the current through inductors will increase.

As you have said the situation is much more complicated in real circuits where different types of component are connected in various ways.

But if you look at the other options, the voltage regulator should prevent these effects unless the RPM is very much lower than standard.

Blimey Keith!

You must be bored. ;)

What is a suction peak in terms of principles of flight.

Obviously I'm missing something, but hopefully some one here can help me out.

Question 1:
Given that the characteristics of a three engine turbojet aeroplane are as follows:
Trust = 50000 N per engine
g= 10 m/s²
Drag = 72569 N
Minimum gross gradient (2nd segment) = 2.7%
The maximum take-off mass under segment 2 conditions in the net take-off flight path conditions is:

a. 101596 kg
b. 286781 kg
c. 74064 kg
d. 209064 kg

Question 2:
The determination of the maximum mass on brake release, of a certified turbojet aeroplane with 5°, 15° and 25° flaps angles on take-off, leads to the following values, with wind:

Flap angle: 5° 15° 25°
Runway limitation: 66000 69500 71500
2nd segment slope limitation: 72200 69000 61800

Wind correction: Head wind: + 120 kg per kt OR tail wind: -360 kg per kt
Given that the tail wind component is equal to 5 kt, the maximum mass on brake release and corresponding flap angle will be:

a. 67700/15°
b. 69000/15°
c. 72200/5°
d. 69700/25°

Question 3:
On a segment of the take-off flight path an obstacle requires a minimum gradient of climb of 2.6% in order to provide an adequate margin of safe clearance. At a mass of 110000 kg the gradient of climb is 2.8%. For the same power and assuming that the angle of climb varies inversely with, at what maximum mass will the aeroplane be able to to achieve the minimum gradient?

a. 121310 kg
b. 106425 kg
c. 118455 kg
d. 102142 kg

It's too late for me to write in my calculations, and for the same reason, please excuse any typos. :bored:

-Anders

Question 1.
The first point to note in addressing this question is that calculation of the maximum take-off mass assumes that a single engine failure has occurred. This means that this three engine aircraft effectively has only two engines.

So total thrust available 2 x 50000N = 100000N.

For small angles of climb, the % climb = 100% x Sin angle of climb

This can be rearranged to give: Sin angle of climb = % climb /100

And Sin angle of climb = (Thrust – drag) / Weight

Combing the two equations above gives:% climb / 100 = (Thrust – Drag) / weight
This can be rearranged to give: Max weight = 100 x (Thrust-Drag) / % climb

Inserting the data provided in the question gives:

Max take-off weight = 100 x (100000 – 72569) / 2.7 = 1015962.963N

This can be converted into Kg by dividing by g = 10 m/s2 to give 101596.2963 Kg or approximately 101596 Kg.


Question 2.
To solve this type of problem it must first be noted that winds do not affect the climb limited take-off mass. The wind correction need therefore be applied only to the runway limit (field limited take-off mass).

The next stage of the solution is to calculate and apply the corrections. the question specifies a correction factor of -360 kg/kt tailwind and an actual tailwind of 5 kts. this gives a correction of -360 Kg/kt x 5kt = - 1800 Kg.

Adding this to the figures provided in the question gives:

Flap angle: 5° 15° 25 °
Runway limitation (kg): 64 200 67 700 69 700
2nd segment slope limitation: 72 200 69 000 61 800

Finally select the flap setting for which the lower of the two limits is greatest. This is 15degree flap which gives a limit of 67 700 Kg.


Question 3.
The maximum climb gradient that an aircraft can achieve is inversely proportional to its mass. This can be stated in the form of an equation:

%climb at new mass x new mass = % climb at old mass x old mass

This can be rearranged to give:

New mass = % climb at old mass x old mass
% climb at new mass

Inserting the data provided in the question gives:

New mass = 2.8% x 110000 Kg = 118461.54 Kg
2.6%

Yes, but of course. Where I went wrong in question 1 was that I was counting with the thrust from all 3 engines. And in question 2, I was making the wind correction to the climb limit also. Thanks for clearing that up for me. However, when it comes to question 3, you have come to the same solution as I, but the book states that the correct answer should be d. 102142 kg.

Cheers
Anders

Which book?

If the aircraft can achieve 2.8% at 110000 kg why should its best climb DECREASE to 2.6% when its weight DECREASE to 102142 kg?

Best climb must INCREASE if weight DECREASES.