OAA Flight Performance & Planning. It's likely to be a bogus answer if you don't agree with it, it wouldn't be the first time...

On another note, I have stumbled upon a couple of questions involving the dumping of fuel to ensure level-off altitude isn't below obstacle clearance altitude during a drift down procedure. Is this common practice, as the only thing I have read about fuel dumping so far is regarding the procedure of doing so prior to landing to ensure not exceeding MSLM?

-Anders

The OAA answer to question 3 is incorrect. Hopefully you should be able to see that is the case by applying common sense. If the aircraft gets heavier it won't climb so well. So if we are happy to be limited to a lower climb gradient we can be heavier.

I'm not an airline pilot, so I cannot comment on what is common practice. But if I knew that keeping any unnecessary fuel on board would cause me to hit the mountain tops, I would certanly dump any fuel that I didn't need.

I agree with you regarding question 3. It's just that as with the case of questions 1 & 2, it's easy to miss corrections, that's why it's seems like a better bet to run it by here, rather than just assume that I'm right and they're wrong.

The fuel dumping questions were:

A:

If the level-off altitude is below the obstacle clearance altitude during a drift down procedure?

a. Fuel jettisoning should be started at the beginning of drift down.
b. The recommended drift down speed should be disregarded and it should be flown at the stall speed plus 10 kt.
c. Fuel jettisoning should be started when the obstacle clearance altitude clearance altitude is reached.
d. The drift down should be flown with flaps in the approach configuration.

B:

During a drift down following engine failure, what would be the correct procedure to follow?

a. Begin fuel jettison immediately, commensurate with having required reserves at destination.
b. Do not commence fuel jettison until en-route obstacles have been cleared.
c. Descend in the approach configuration.
d. Disregard the flight manual and descend at Vs + 10 kts to the destination.

Correct answer for is a. for both questions. Looking at them in hindsight I realize that those are the only reasonable answers, it's just that when I came across them there had been no mention of fuel dumping in the previous chapters that I had read, other than for the purpose of ensuring MSLM, and I suppose that the intention after engine failure is to land as soon as possible, thus fuel dumping should be considered pertinent. However there is no mention in the question of what phase of flight is being referred to, and dumping of fuel doesn't seem to be advisory unless absolutely necessary. What had been brought to attention in the previous chapter however, was the use of drift down profile graphs, where one was to determine if the desired altitude, with regard to obstacle clearance, could be met with the current mass, and if not, (to my understanding) a second graph should be used to determine whether or not vertical clearance could be achieved using horisontal distance instead. Thus the reason for my inquery.

You are however absolutely right, Keith, it definately seems like a better idea, if need be, to dump some fuel rather than run in to a mountain top.

Cheers
Anders

Hi I Am really confused and finding it difficult. Can anyone please help

Question is: indicated altitude =20,000 feet and the temperature is -35*C. What is the true altitude?
The equation given is -

true altitude= indicated altitude + (ISA_Deviation) x Indicated Altitude.
-------------------
(T)(k)






And if somebody can really break down each part of the equation and how it works, it would help immensely.


Question 2)
Indicated altitude 32,500 temperature is -32*. What is true altitude?

If you go back to post number 33 in this thread you will find something very similar to your question.

Could somebody be kind enough to tell me how many questions there are in the EASA ATPL EXAMS for
Principles of flight, meteorology, mass & balance, human performance and limitation, communication?

Has anyone sat the new exams and whAt is the differnce if there is any. More difficult I.e. complex questions?

Keith William thank you for your input on the previous post. It was very helpfull.

Hey again...

On a ground pressurisation test, if the cabin suffers a rapid depressurisation:

A. The temperature will rise suddenly.
B. Water precipitation will occur.
C. Damage to the hull may occur.
D. Duct relief valve may jam open.

I had originally marked answer C, but according to the solution it should be B. Is this correct, and if so, would anyone care to elaborate?

-Anders

The sudden pressure drop will cause a sudden temperature drop.

If there is sufficient moisture in the air, and if the temperature falls below the dew point, it will cause water droplets to precipitate out of the air.

The unstated assumptions in this question are that:

a. There is sufficient moisture in the air.
b. The pressure and temperature drops are sufficiently large.

So the use of the words "will occur" in option C is probably a bit too strong. "May occur" would be more accurate.

But the other options are all wrong, so option C is the best answer.

Thanks Keith.

-A

ATPL Theory study
 

Just embarked on the Bristol GS study, any tips on helping me ease my pain?:{

Have read many of the posts in this forum and see a lot of people are struggling with the Gen Nav side of things, I've just purchased a book from Baz at Bristol GS 'Mathematics for Aviation', I sure hope that helps me understand the subject a lot better when I receive it :ok:

Hey,

The tendency to call to mind common experiences or scenarios from the past and link them incorrectly to a perceived mental model is called:

a. Confirmation bias.
b. Action slip.
c. Environmental capture.
d. Frequency bias.

Supposedly the right answer should be d.
Is this correct? Couldn’t seem to find anything about frequency bias in reference to HP. I suppose that if one breaks it down, then it could be the right answer. However the definition of environmental capture seems quite fitting in the context…

-Anders

Question about weather radar of airbus a321!!
 

Why on A321 the weather radar system has two function (auto and manual)? What difference between those function and which function more reliable?

Are you serious?..stupid question....
 

Question: Which combination of answers of the following parameters give an increase or decrease of the take off ground run: 1 decreasing take off mass 2 increasing take off mass 3 increasing density 4 decreasing density 5 increasing flap setting 6 decreasing flap setting 7 increasing pressure altitude 8 decreasing pressure altitude

A.2, 3, 6 and 7

B.1, 3, 5 and 8

C.2, 4, 5 and 7

D.1, 4, 6 and 8

I think it would have to be B because it's the only one which lists conditions which all affect ground run in the same direction, as in, they all cause it to decrease. All the rest contain some conditions which cause and increase and some which cause a decrease.

Pretty bad wording all the same.

Yea correct answer is B!!

new formulae
 

hi dear just try this ZFW=TRAFFIC LOAD+DOW 112500=traffic load+80400 traffic load=11250-80400 =32100 :ok:

W/V Calculation
 

Hi

For a given track the:

Wind component = +45 kt
Drift angle = 15 left
TAS = 240 kt

What is the wind component on the reverse track?

a) -55 kt
b) -65 kt <-- Marked Correct
c) -45 kt
d) -35 kt

With an E6B I get around -55. It probably uses the formula Wind Speed x Cos (Wind Direction - Course).

Is it normal/possible to have a 10kt difference if done with a CRP-5 or I am missing out something?

Thanks

Are you accounting for decrease in TAS to ETAS due to large drift angle at high TAS?

can you show please how you are solving it. thanks

To solve this type of problem using the CRP5:

a. No heading is specified so select one at random, 000 degrees for example.

b. Set the high-speed slide with the centre dot at 240 kts.

c. The question specifies a +45 wind component. This means a 45 kt tailwind component. This means that the ground speed is 240 TAS + 45 kts = 285 kts.

d. The question also specifies 150 left drift. So draw a cross where the 15 degrees left drift line crosses the 285 kt arc.

e. With 15 degrees left drift the track is 345 degrees. So the reverse track is 165 degrees.

f. Rotate window to align 165 degrees with true heading index.

g. The cross now indicates 19 degrees right drift. Align the 165 degrees track with 19 degrees right drift. The cross now indicates 15 degrees right drift. Align 165 degrees with 15 degrees right drift. The cross now indicates 16 degrees right drift. Align 165 degrees with 16 degrees right drift. Both the cross now indicates Move and the outer scale now indicate 16 degrees right drift, so the drift is balanced.

h. The cross is now on an arc 65 kts below the centre dot. The wind component was a headwind during the outward leg so it must be a tailwind during the return leg This means that the wind component is -65 kts.


There are quicker ways of doing the job (I saw one once here in pprune but I can't remember what it was), but the above is the method described in the CRP 5 instruction sheet.

Thanks Keith

Hdg 360
CRS 345
TAS 240
GS 285

W/V comes out to be 116/82

With CRS 165 and wind 116/82 the headwind component comes out around -53.8 with this formula: Wind Speed x Cos (Wind Direction - Course)

Shouldn't the answers be close together if not exactly the same?

Regards

I think that your wind should be closer to 112/82

And I think that your equation will get you wind component down the track.

If you use Windspeed x Cos(Wind direction - Heading) You get

82 x Cos(112 - 149) = 65.5

ok, I think more than the wind its the formula that's different

I was using Wind Speed x Cos (Wind Direction - Course)

Whereas you have mentioned it is:

Windspeed x Cos (Wind Direction - Heading)

Is (Wind Direction - Course) wrong or is it used somewhere else, I think i read it somewhere :confused:

Looking again at my description of the solution using a CRP5, I can see that I have made an error in that I assumed that the 240 knot TAS plus the 45 knot wind component would give a ground speed of 285 knots. This would be true if there was no drift, but it is not true with 15 degrees drift, because drift adds a lateral component to the ground speed.

If we split the TAS into one component along the 15 degree drift track, and another component at right angles to it we get 231.8 knots along the track and 62.2 knots across the track. Adding the 45 knot wind component to the along-track speed gives a ground speed of 231.8 + 45 = 276.8 knots.

The modified solution using the CRP5 become:

a. No heading is specified so select one at random, 000 degrees for example.

b. Set the high-speed slide with the centre dot at 240 kts.

c. The ground speed is 276.8 knots.

d. The question also specifies 150 left drift. So draw a cross where the 15 degrees left drift line crosses the 276.8 kt arc.

e. With 15 degrees left drift the track is 345 degrees. So the reverse track is 165 degrees.

f. Rotate window to align 165 degrees with true heading index.

g. The cross now indicates 19 degrees right drift. Align the 165 degrees track with 19 degrees right drift. The cross now indicates 15 degrees right drift. Align 165 degrees with 15 degrees right drift. The cross now indicates 16 degrees right drift. Align 165 degrees with 16 degrees right drift. Both the cross now indicates Move and the outer scale now indicate 16 degrees right drift, so the drift is balanced.

h. The cross is now on an arc at approximately 55 kts below the centre dot. The wind component was a headwind during the outward leg so it must be a tailwind during the return leg This means that the wind component is -55 kts.


This matches the solution that you got with your E6B.

It is curious that my error led me to answer that was marked as being correct. This makes me suspect that the author of this question made the same error.

Thanks Keith

I think your previous calculations were right. Perhaps I was mixing up two different modes of the E6B.

1) To find the W/V

Hdg 360
CRS 345
TAS 240
GS 285

W/V comes out to be 116/82


2) To find HDG and GS on reciprocal course

W/V 116/82
CRS 165
TAS 240

GS comes out to be 178
HDG comes out to be 150

Thus the wind component comes out to be TAS-GS (240-178) = 62kts Headwind.


3) There is another mode that calculates the X-wind and H-wind

For a W/V of 116/82

If you work it out with Course i.e. 165; H-wind component comes out to be -54 and X-Wind is 62 from the left.

If you work it out with Heading i.e. 150; H-wind component comes out to be -68 and X-Wind is 46 from the left.

Since we are maintaining the track and not letting the aircraft drift, we'll be heading into the wind so more headwind component. If we are not heading into the wind then less headwind component but more x-wind and the aircraft will drift.

I was using the course instead of headwind thats why the answers were different.:ugh:

This is a somewhat lengthy explanation (and using different values) which confirms Keith’s original solution.

You have only been given a fixed numerical value for the TAS, so all other numerical values for the Outbound Track must be based on assumption.

That’s not as ambiguous as it seems at first glance because you are told that the effective Wind Component is +45 kts, making the GS 240 + 45 = 285 kts.

You are also told that the Drift is -15°. Working on the basis that Drift is invariably FROM Heading TO Track, choose some simple numbers to make your calculations easier: I used Heading = 115°T and Track = 100°T.

With these four components we can now establish the Wind Vector.

On the CRP-5 (HI SPEED):

Place centre dot over TAS = 240 kts

Rotate central bezel to place HEADING = 115°T on inner scale under TRUE HEADING index on outer scale

Make a Wind Mark where the straight, vertical Drift line (=15° LEFT) crosses the curved, horizontal speed line at 285 kts

Rotate the central bezel to place the Wind Mark on the central Drift line UNDER the TAS

Read the Wind Direction under the TRUE HEADING index = 230°

Read the Wind Speed by counting down to the Wind Mark from the TAS = 80 kts

Wind Vector = 230° / 80 kts

To calculate the inbound effective wind component, we note that we have a W/V of 230/80, a TAS of 240 kts and a reciprocal Track of 280°T (100°T + 180°)

On the CRP-5 (HI SPEED):

Place centre dot over TAS = 240 kts

Rotate central bezel to place Wind Direction = 230°T on inner scale under TRUE HEADING index on outer scale

Make a Wind Mark 80 kts BELOW the TAS at 160 kts (240 – 80)

Place Track = 280°T on inner scale under TRUE HEADING index on outer scale

Note Wind Mark has moved to indicate 18° of Right Drift

Align 280°T on inner scale with Drift = 18°L on outer scale

Align Track = 280°T on inner scale with Drift = 18°L on outer scale

Note Wind Mark has moved to indicate 14° of Right Drift

Align Track = 280°T on inner scale with Drift = 14°L on outer scale

Note Wind Mark has moved to indicate 15° of Right Drift

Align Track = 280°T on inner scale with Drift = 15°L on outer scale

Note Wind Mark still indicates 15° of Right Drift

The Drift on the outer scale now balances the Drift on the Drift grid, meaning the computation is complete

Read Heading = 265T under the TRUE HEADING Index
Read Ground Speed = 177 kts (ish) on curved speed line

Difference between TAS and GS = 240 kts – 177 kts = 63 kts

GS is slower than TAS, therefore, effective wind component = -63 kts

Closest answer = -65 kts


NOTE that it you are completing these questions under the EASA Learning Objectives, you are required you to use a Navigation Computer and NOT trigonometry. In fact, there ARE some CQB questions where the mathematically more precise trigonometrical solution has been provided as an incorrect option. Bonkers maybe but don't shoot me, I am merely the messenger, I get enough flak for the examiner in class off the students!

Thanks BGS

Performance: Domain
 

Hi there,
I'd need some feedback regarding the following question - I believe that the answer provided is incorrect.

QDB says that the correct answer is 277.5 m but I come up with 279.7 m applying the following formula:

Domain 1/2 width = 60m + 1/2 wing span + 0.125 D
Domain 1/2 width = 60 + 32.2m + (0.125 x 1500m)
Domain 1/2 width = 279.7m

Thanks

Hi

For aircraft with a wing span of more than 60m*domain starts with a semi-width of 90m.*

So it will be 90 + (1500 x 0.125)

thanks for the prompt respond, much appreciated! I guess I was wrong in this case.

if a radio altimeter fails
 

the right answer is actually a) height information disappears

I know the question is tricky and too type specific, so the only answer which can't be proved wrong is a.

cheers:ok:

Converging aircraft
 

Hi

Can anyone tell me if there is a formula for when 2 aircraft leave different points at different speeds. when will they meet if the distance between the 2 is a certain number of miles?

Assuming that they both take-off at the same time:

Flight Time to meeting point = Distance / (VA + VB)

Where

Distance is distance between the two starting points.
VA is ground speed of aircraft A.
VB is groudn speed of aircraft B.

Position of meeting relative to start point A = Time x VA

Position of meeting relative to start point B = Time x VB

Assuming that A and B and the meeting point are in a straight line and that the meeting point lies on the line between A and B

Hey guys,

Just came across this question:

Where are easterly and westerly jets found?
a. Northern hemisphere only.
b. Southern hemisphere only.
c.Northern and southern hemisphere.
d. There are no easterly jets.

The correct answer is a, but I can't seem to find anything to confirm this. Supposedly easterly jets only occur in the northern hemisphere and somewhat rarely(?) at that. I haven't however been able to locate any text stating that this is the fact. I have on the other hand found some indications that easterly jets do in fact occur in the southern hemisphere. Anyone able to shed some light?

You are of course correct Dick, but Bradders 147 appeared to be looking for something simple, so I went for the most simple scenario.

If there is a simple equation that covers all possible combinations of track geometries I don't know what it is. Do you?

If we restrict ourselves to the "classic" sub-tropical jets then they are westerlies because the thermal equator is more or less at the geographic equator - except in one case. In the northern hemisphere summer, July, August or thereabouts, the tremendous heating of the Tibetan plateau drags the thermal equator well north of the geographic. Now the upper air flow out of the upper high is subject to geostrophic forces opposite to normal and the winds turn easterly

This upper easterly flow can reach jet speeds right up near the trop in the area between N India and W Africa. You can find them on the upper winds/temps charts or sometimes on regional sig wx chats

There are other forms of "jet", for example the low level cold front jets that occur equally in both hemispheres

Performance - Landing (Class A)
 

Hi all,
just came across this (apparently) weird question, can someone explain how to figure out the correct answer or am I missing something?

At maximum landing mass, the structure of the aircraft is designed for a rate of descent:
- 250 fpm
- 600 fpm
- 200 fpm (this would be the correct answer, but how is this figured out, is there a formula or something :confused:)
- 220 fpm

Thanks in advance!

And I need a brake now!

Its a design function in the relevant certifying specs

Hi all,
I have a question about easterly waves. Since in summer prevalent upper winds in area above ITCZ are westerlies, and movement of storms is dictated by upper winds, is the westward movement of easterly waves and corresponding storms consequence of Tropical Easterly jet?