I definitely don`t wanna argue, but if you say this is true, then it is just another junk question, which has nothing simmilar with the student`s knowledge. Just a stupit guessing, which way is the correct way. But we all know, that there are many questions like this in ATPL Theory, so I just have to believe your calculations.. Good luck to the students getting this kind of question at the exam..

Yes some of the questions make you wonder and FP is much better than most.

As stated this question does cause issues and is one I always explain as discussed in previous posts. It must be right as some students are getting 100% with this type of problem included!

Given that the characteristics of a three engine turbojet aeroplane are as follows: Thrust = 50 000 Newton / Engine, g = 10 m/s², Drag = 72 569 N, Minimum gross gradient (2nd segment) = 2.7% SIN(Angle of climb) = (Thrust- Drag) / Weight. The maximum take-off mass under 2nd segment conditions is:

The only way i get the answer that the data base gives is by assuming one engine is out...but at no point in the question does it say that it is.

perhaps you only need the segments when the engine quits......

First, second and third segment weights are based on having one engine inop and still meeting the required gradient.

Yes, I also noticed when learning for my ATPL, that this kind of questions is always based at one engine inop situation. Pay attention on this and good luck

Two performance questions that I have found question banks with different answer.


On a twin engined piston aircraft with variable pitch propellers, for a given mass and altitude, the minimum drag speed is 125 kt, and the holding speed is 95 its. The best rate of climb speed will be obtained for a speed of:

One question bank says 95, the other 125.

What other speeds can you derive from the above question?

The other question I found is for a piston engine maximum endurance is achieved at:

One bank says maximum rate of climb speed, the other at maximum lift to drag ratio, one and the same?

Hi

Q.1. During a flight to Europe, scheduled in MNPS (Minimum Navigation Performance Specification) airspace, you expect to cross the 30oW meridian at 2330 UTC; you will normally be:

a) in a day flight route system
b) in random airspace
c) in a night flight route system
d) out of the organised route system

why not (d)?

Q.2. During a flight to Europe, planning in MNPS (Minimum Navigation Performance Specification) airspace, you expect to cross the 30oW meridian at 00H30 UTC, you will then normally be:

a) within the organised night-time flight track system
b) with the organised daytime flight track system
c) out of the organised flight track system
d) in a random space

why not (d)? (unless you want to rule out "space" instead of "airspace")

thanks

OTS times are:-

East or night time - 0100 to 0800 UTC at 30w
West or day time - 1130 to 1900 UTC at 30w

Flying an OTS route is optional and a flight not entirely following an OTS track would be called a random flight/route as are flights outside the time periods.

There is no such thing as 'random airspace'. Where did you get these questions? I thought they were corrected years ago.

Both flights are operating outside the standard OTS times and are random flights. Q1 = D, Q2 = C (as the better answers).

Thanks Richard

They still stand uncorrected in some QBs among many others.

Regards

Anyone seen this question before? It was in the Performance Exam today and though I cant recall it verbatim, it was regarding the benefits of reduced thrust.

Out of the 4 options only 2 were viable,

1) Reduces trip fuel
2) Reduces a limited VMCG.

I can see arguments for both answers, one side you do use less fuel when you use reduced thrust, VMCG could be lower since theres less thrust. Any suggestions?

Re yesterday's Performance exam I actually thought the answer to the above question was the option involving Field limiting take off weight being a lot more/less (or whatever...) than Climb limiting. I think it may have been answer C. Either that or reduce a limiting VMCG...

As far as I could see...

Reduced thrust CAN be used on a wet runway as long as it is not considered contaminated. So it couldn't be that...

Trip fuel option just sounds wrong because that is not a reason to use it...the saving would probably be small and I have never seen that given as a reason to reduce take off thrust...

It will reduce the VMCG but as far as I'm aware there is no data to support that in the CAP 698 beyond the VMCG table for max thrust. I have heard of it being used as a technique in the real world but I'm just not sure whether it would be considered the right answer for the purposes of the exam...

In hind sight I would have gone with the VMC option only because you can't count on the use of reduced thrust in case of windshear etc. maybe someone appealed it.

What did you put for drag corffiecent question? I put lowest point which question bank people say is correct but app examiner is looking for tangent answer. We appealed that one.

Don't remember that one unfortunately, do you mean the one with the graph in the annex?

Yes thats the one

Use of reduced thrust for climb increases total trip fuel and should be evaluated by each operator.... Takes more time to reach the optimum level

See we use reduced thrust is replaced with climb thrust at 1500 so saves fuel up to that point as well as other benefits engine wear etc

Can you remember more specifically what the question was asking? I can't remember what point it was asking to identify on the graph?

ATPL question help
 

Dear,

I know i'm not in the good thread but i need an answer :

Question 16- REF 330076

(Refer to route manual Chart Nap) date 1999
From Reykjavik ( 64°10 ' N - 020 °N 020° 00'W) to Amsterdam ( 52°32'N 004°50'E), what is the distance ?

I don't understand how they found 18° of Lat difference... Can someone explain me how ? (In the correction they found 18°-Bristol)

Thanks

Alan,

Many readers do not have access the material to which youare referring. You might get a betterresponse if you were to post a copy of the question and the explanation to enableeveryone to examine it.

From what you have posted I can see that there is one errorin that the longitude of Reykjavik should be 022W and not 020W.

Alew, I agree with Keith. Plus PLEASE I think you meant to say too!

However if the question is just asking for a distance then you should just draw a straight line on the chart and using dividers/compasses/rule and the fact 1 degree of latitude = 60 nm you should get around 1080 nm. I do using both the NAT charts both using dividers and the appropriate chart scale and an inch ruler.

So on information provided don't know where the 18 lat difference comes from, sounds rubbish to me.

ATPL question :)
 

Hey guys im stuck here, could you please please help me out with the question,
if possible with explanation how to do it too.

You are flying along the 315* radial from a VOR/DME. On a heading of 310*M and flying a TAS of 450 kt, at 0650 UTC you are 93NM from the DME. At 0655 UTC you hav reach 135 NM from the DME. Variation off the map is 16*E. What is the most likely W/V?

answers
A 005/54
B 165/70
C 105/65
D 180/70 probably correct

This is what you have:

TAS 450 kts
Groundspeed 504 kts
Drift 5 degrees R

Pick up a flight computer and it will reveal the W/V.

I don't have mine any more (gave up teaching this stuff 18 months ago) but answer D indeed looks a likely candidate.

Hello. I have done some calculations. My Result is a little bit different from all answers.

In the first stage, you disregard variation, because VOR radials are magnetic bearings and your heading is also magnetic. So the angle of heading 5 degrees to the left from your track. Your GS can be obtained from time and DME: 5 minutes = 42 NM -> 1 minute = 42/5 = 8.4 -> 60 minutes = 504 KTS.

From here you work with a Cosine sentence:
c^2=a^2+b^2-2*a*b*cos(Gamma)

you fill in:
a:450
b:504
Gamma: 5 degrees

you get C = 68 kts - wind velocity.

if you have this, you use Sine sentence:
a/sine(Alpha) = b/ sine(Beta)

you fill in:
a: 68
b: 450
Alpha: 5 degrees

Then you get angle of 35 degrees.
This is the angle, which is added to your track, from where the wind is blowing. So, if you add 35 to 315, you get wind blowing to 350, from 170.
Wind is 170/68 Magnetic, if you add variation, you get final wind blowing 186/68.

Please, someone, correct me if I`m wrong.

As Lightning Mate said pick up a flight computer and it's a lot easier than cosines and sines.

To work out W/V you need heading, track/course, TAS & GS - basic triangle of velocities stuff just asked in a more practical manner.

GS = 42 NM in 5 mins = 504 kts.

In this case as both heading and track were magnetic work in magnetic on the CRP-5 and get a magnetic wind of 170/70 then ADD the variation back in to get true W/V 186/70 so D.

If you have a mixture of true and magnetic I would always work in true.

Gross error check - wind blows from heading to track so W/V must have a southerly element in this case, rules out answer A to start with.

Why do some people make the solution to a simple problem so difficult ? :ugh::ugh:

Principle of Flight
 

Hi Guys,

can anyone recommend me a good free online source to study for the Principle of Flight exam?

Thanks :)

Try your approved school - that's what you are paying them for.

Some while ago I worked for the CAA on the PoF CQB.

I have a whole collection of questions.

I am doing the atplonline QB at the moment and my distance learning provider has not the best material online. I thought someone can point me in the right direction. Preferable key facts especially about high speed aerodynamics.

Pittslover,

While not directly satisfying your request I will nevertheless recommend HH Hurt's classic Aerodynamics for Naval Aviators. The FAA hosts a PDF copy, not under US domestic copyright protection, here: http://www.faa.gov/regulations_polic.../00-80T-80.pdf

Although reading this from cover to cover will be overkill for ATPL PoF exam purposes, it is sufficiently simple to supplement learning without requiring much prior knowledge.

met doubt : explanation for question
 

hi, i have a question from met, I want to know if the explanation is correct.
if an aircraft in northern hemisphere flies from high to low, it will experience _________ drift
ans. starboard

explanation: if winds are from left, then it will experience right drift, it is because aircraft is flying towards ( or in the) low pressure, so the winds will be from left. hence starboard drift.
rule applied is coriolis force.

Do a search for Ballots law.

All fluids tend to flow from areas of high pressure to areas of lower pressure. But because of the rotation of the Earth, Coriolis effect causes the wind to circulate around areas of high and low pressure. The direction of this circulation is depends on the location (northern or southern hemisphere), and whether the area is one of high or low pressure. The outcome of these effects is predicted by Buys Ballot’s Law, which states that when standing with one’s back to the wind, the low pressure area is on your left.

If an aeroplane is subjected to a crosswind it will drift in the direction of that wind. In this question the aeroplane is drifting to the right, so the wind must be coming from its left side. Applying Buys Ballot’s law this means that the aeroplane in this question must be flying towards the area of low pressure.

If you need a more detailed explanation you should read up on coriolis effect, which is part of the ATPL Met syllabus.

thank you!

If in doubt, draw it out! :)

yes, thank you very much! :)

From memory Tx means braking action...

Polar Stereographic Chart
 

HI

Can anyone please tell me if questions on the Polar Stereographic chart are still being asked in the ATPL Gen Nav exam?

Hearing that it has been taken out of the syllabus?

Cheers

Yes they are still there in GN and questions are asked.

Thanks for this Richard..

Cheers