ATPL theory questions

[dartagnan]

found this question somewhere...

[Mercenary Pilot]

I think TX is maximum temp, and TN is minimum temp forecast. :8

[HeliCraig]

Never seen it actually used (at least in the UK), but it is Max Temp.

This page is always good: http://aviation.weathersa.co.za/codesexpl.php

[LH2]

Never seen it actually used (at least in the UK)

Commonly seen at a number of Mediterranean airfields, e.g.:

TAF LEBL 261700Z 270024 34010KT CAVOK
TEMPO 0024 FEW025 PROB30
TEMPO 0010 34015G25KT
BECMG 1214 20010KT
BECMG 1820 35010KT TX16/13Z TN08/06Z

[shlittlenellie]

Mostly seen in Spanish met.

[Willows]

Not in the abbreviations list on the Met office site. I shall pen it in on my copy.

Not sure if these links will work. You need a login (it is free)

abbreviations (http://secure.metoffice.com/aviation/abbreviations.jsp)

TAF decode (http://secure.metoffice.com/aviation/taf_decode.jsp)

METAR decode (http://secure.metoffice.com/aviation/metar_decode.jsp)

[RingwaySam]

Hi all,

I came across this one a few days ago, although it's todays, what does the RS0109KT 050209KT 230109KT Stand for?

LPMA 270000Z 03009G20KT 350V070 9999 SCT018 19/12 Q1023 RS0109KT 050209KT 230109KT
LPMA 262300Z 270009 03010KT 9999 SCT018
TEMPO 0009 03012G22KT SCT018 BKN030

Thanks!

[pilotms]

These are some points were they are reporting the Winds. RS stands for Rosario.

[Bus14]

LPMA is Funchal/Madeira. RS (Rosario) is about a mile short of the 05 threshold and sits at the end of a valley down which winds blow which can cause all sorts of chaos on short finals. 05 is the touch down anemometer for RWY05 and 23 is the same for RWY23. The main wind is the tower wind. The Portugese AIP has the detailed rules on the limits that apply for take-off and landing.

Although the runway is now quite long, LPMA remains one of the more challenging/fun airports to visit. Just think of it as an aircraft carrier moored alongside a big cliff and you get the idea:cool:.

[RingwaySam]

Interesting. Thanks! :ok:

[PintofT]

Not that I'm work-shy, you understand......but is one expected to remember each of the Chicago Convention's articles & annexes, what each refers to and their implications?

And I thought the FREDA thing was tough!! :ooh:

[Whirlygig]

You might one, maybe two if you're unlucky, questions asking which Annex number deals with .... whatever.

However, you are expected to know the contents and their implications (if not the number).

Cheers

Whirls

[greekboy]

Table 20 ICAO Annexes
Annex
Subject
1
Personnel Licensing (Getting a license is my Number 1 priority)
2
Rules of the Air (2 Sets of Rules, VFR and IFR)
3
Meteorological Services (3ºC/100ft DALR)
4
Aeronautical Charts (4 Cardinal Points)
5
Dimensional Units (CRP 5)
6
Operation of Aircraft (DC6)
7
Nationality and Registration Marks (The League of Seven Nations)
8
Airworthiness (Looks like a propeller)
9
Facilitation (NEIN in German – Immigration)
10
Aeronautical Communications (100 for the Operator)
11
Air Traffic Control Services (1 to 1 Personal Services)
12
Search & Rescue (The one before Accident Investigation)
13
Accident Investigation (Unlucky for some)
14
Aerodromes (14 Aerodromes around Heathrow)
15
Aeronautical Information Services (Looks like IS)
16
Environmental Protection (16 Age of Consent, use protection)
17
Security (17ft security fence required)
18
Dangerous Goods (At 18 you can drink but it’s DANGEROUS to drive)

[greekboy]

Oh by the way. this is not my brainchild but some earlier student from BGS.

g luck

gb

[PintofT]

Thanks for that guys!

[Air1980]

Any other memory aids? Like for example the chart specifications (Lambert etc)?

[PintofT]

Air1980, PM sent. May be of help.

[Mohit_C]

greekboy one correction. The DALR is 3ºC/1000ft OR 1ºC/100m.

[thibautg78]

Hi all,

Here are a few questions from oxford school regarding instrumenation which, actually, are not that hard but i prefer to ask for advise before submitting my assessment.
So enjoy it !

1) With a fully equipped FMS aircraft the following selections are usually utilised in flight :
a- LNAV only
b- LNAV and VNAV
c- VNAV only
d- LNAV and VNAV but not simultaneously

I'd answer d), cause if i have well understood lnav is used during cruise while vnav is more concerned with climb&descent, so the use of both modes cannot be simultaneous. Is that right?

2) If the radio altimeter fails :
a- height information disappears
b- aural warning given
c- radio alt flag, red lamp, and aural warning given
d- radio alt flag and red lamp activates

I'm quite hesitating between c) and d).

3) If an aircraft fitted with two FMSs, the pilots' displays show contradictory information. What mode of FMS is in use?
a- independant mode
b- dual mode
c- single mode
d- separate mode

Answer a) ?

Thanks guys!
:ok:

[_ShIfTy_]

I would go with answer B to the first question. L-NAV and V-NAV are both used during most part of the flight.

[AlpineSkier]

" I would of went..."

Is it only when they can't tie their shoelaces that they are officially morons ?

[hvogt]

thibautg78,

I can't say I have always been successful in my flight school's quality checks, but still I'd like to take an educated guess.

LNAV and VNAV can be used simultaneously, and in fact they are during climb and descent. How else would you follow an SID or a STAR? However, VNAV is not usually used during cruise flight. Instead the APFDS is used in height lock, but this does not mean you cannot operate LNAV and VNAV together. So, the correct answer is "b".
I'm not certain either. In an A320 an RA failure gives a single chime (i. e. an aural warning) and a master caution light (not a red lamp, as in the question, but still a lamp). Hence, I would tend to "c".
The correct answer is "a" (independent mode). In single mode only one FMS is operational. I've never heard of dual or seperate mode.
Good luck with your assessment.

[thibautg78]

Thanks guys, i'll keep you aware of the right answers once i have submittted my on-line assessment.
;)

[mad_jock]

I think in the majority of old types the first thing you will know about a rad alt failure is the EGPWS going to the wind. Then when you manage to find out where they have fitted the sodding thing on this airframe it will have a red flag on it.

But a word of caution the ATPL's use the 737 as it's generic Jet.

Be a bit careful using Airbus as a reference to answer questions.

On my type the answers would be

D A A

[Coldbear]

hi,

On the A320 the answers would go like this:

1) B - We can and do use LNAV/VNAV together.

2) B - it's an amber light on the airbus

3) A

hope it helps and good luck,

[Piltdown Man]

1. In the UK, HDG is the normal selection. (But I think they are looking for B). FWIW, the use of VNAV on its own is generally not recommended. And on an Embraer E jet, I wouldn't use VNAV at all because it's pants.
2. What a crap question, it is too type and version specific.
3. Again, a type specific question, but A would probably be the best answer.

If I was asked these questions in an exam, I would refuse too answer them as they are too type and version specific.

[giladal]

HI guys,

Ok, here is the question. I understand how to get the NDB true direction, but for the love of god, i cant understand how they came up with 273 Radial as the one the aircraft has intercepted.. Anyone?


An aircraft maintaining heading 140°(M) at 057°E (VAR 12°E) takes a relative bearing reading of 295° from an NDB station co-located with a VOR at 32°N 088°E (VAR 8°E). Find the true direction of the straight line that should be plotted from the NDB to the aircraft on a Mercator chart. Which radial has the aircraft intercepted at that moment? (Use mean latitude 27°N).

Correct answer: R-273°, 274°

[giladal]

Wow.. 100 + views and no replies...

No worries, I figured it out..

You take the True bearing, add the conversion angle and then convert it to magnetic.

274T + 7 - 8E = 273 Radial.


Now I am stuck with Lambert charts...

[RichardH]

Good question, the key bit is understanding for NDBs variation is measured at the aircraft and VORs measured at the VOR. Also straight line on Merc = Rhumb Line. Remember radio bearings follow great circles.

NDB - Hdg(T) = 152 + 295 = 447 -360 = 087(T) from A/C to NDB
As there is more than 2degrees ch long need to apply convergency.

CGY = CH Long (31) x sine 27 (.4540) = 14
Conversation Angle = 14/2 = 7
RL Track from A/C to NDB/VOR = 094 + 180 = 274 to plot on Merc from NDB.

GC Track at NDB/VOR = 087 + CGY (14) = 101 + 180 = 281(T)
281 - 8E at VOR = 273(M) = radial or QDR

Might help to draw a picture and there are a number of ways to do this question.

Ah happy days!

[giladal]

Thank you very much richard. I really appreciate it :)

Here is another problem I got stuck with.. Any idea?

Here is my question:

Following a Rhumb line track starting from E in Lat N050.

E to F 000T, 300NM
F to G 090T, 300NM
G to H 180T, 300NM

What is the Rhumb line bearing and distance of H to E?

The RLB just at a glance seems to be 090... but the distance??


I tried and tried, and then skipped, but would really appreciate it if someone can help.


Thanks...

[G CEXO]

466.7 NM

that's the best I could come up with.



By the way, you will need to use the Departure formula as you are on a latitude of 50 degrees. Hence 60 NM is not 60 minutes.


Departure = change in longitude x Cos Latitude

Change in Longitude in minutes.

[giladal]

Hi G.

Thanks for trying, but the answer is 336Nm.

300 / cos55 = 523nm

523 x cos50 = 336Nm

Thanks again ;)

[RichardH]

Pleased to assist.

Assume E is at N50 (as given) on prime meridian.
E to F 300 so F is N55 E/W000
F to G 300/cos55/60 = CH long of 8:43 so G N55 E008 43
South to H will bring you to N50 E008 43

I think RL bearing is 270 (but I am a little tired).
8:43 = 523 min * cos 50 = 336 nm.

You must get into drawing diagrams.
Very common type of question and is all about departure and ch long varying with cosine of latitude due to spherical earth.

Members of the flat earth society think you end you back at the start!

[giladal]

Hey Richard,

Ok., here is a new one ;)


This one seems a little bit more complex.

Two aircraft are converging to the same reporting point at FL 330. The wind is 350 / 60kts. At 0700 the distance between the aircraft is 130NM, and the controller requests aircraft A to reduce the Mach number so there will be 3 minutes separation over the reporting point.

Aircraft A: .76M, TAT = -21c, Track 330(T)
Aircraft B: .79, TAT = -19c, Track 060(T)

1)What is the ETA of aircraft A at the reporting point?
2) What Mach number should aircraft A maintain if the SID temp is -45c


Looking forward to hearing from you.

[AVIATROZ]

I need help regarding solving of these questions,i am not able to get the concept to solve these meteorology problem,i would really appreciate if some one could help me out in these concepts

Q1. MSA given as 12000ft, flying over mountains in temperatures +9*c ,QNH set as 1023.what will the true altitude when 12000 ft is reached?
a. 11,940
b. 11,148
c. 12,210
D. 12,864
ANS -D{..........how plz explain.....}

Q2.Flying at FL135above the sea, the Radio altimeter indicates a true altitude of 13500ft. the local QNH is 1019hpa.hence the crossed air mass is on average,

a. At ISAstandard atmospher
b. Colder than ISA
C. Warmer than ISA
D. There is insufficient information
ANS-B{...why plz explain...}

Q3. You are flying at FL160 with an OAT OF -27*C.QNH is 1003 hpa. what is your true altitude?
a.15540 ft
b.15090 ft
c.16330 ft
d.15730 ft
ANS-B{.....how plz explain.....}

Q4. What is approximate vertical interval which is equal to a pressure change of 1hpa at an altitude of 5500m?
a. 8m{27ft}
b. 32m{105ft}
c. 64m{210ft}
d. 15m{50ft}
ANS-D{....how plz explain...}.

thanks

[Conan the Vulgarian]

Q1. True altitude differs from indicated by 4% for every 10 degrees difference in temperature from ISA. True altitude is higher than indicated in warm air, lower in cold air. At 12,000 ft ISA temperature is -9 degrees, if the actual temperature is +9 then the temperature is ISA+18, so 1.8 x 4 = 7.2 percent difference, the actual altitude is 107.2% of the indicated.

Q2. If the QNH was 1013mb, you'd be at 13,500 feet when at FL135. Since if you increased the subscale setting to 1019 the indication would increase, then FL135 is actually a bit lower than 13,500 ft. (You're flying into higher pressure air than ISA, so you're higher than you would be if it were ISA). Since the radio altimeter says you are really at 13,500 feet, you must be in warmer than ISA air.

Q3. FL160 is 15,730 feet above sea level if the QNH is 1003. (FL measures from 1013 and since the pressure at sea level is 1003 the datum you are measuring from is below sea level). OAT is -27 wheras ISA temperature at your level would be -17, so it's ISA-10, so take off 4% to get 15100ft - 15090 is close enough.

Q4. 50ft and you just have to memorise these, although you could make a decent guess if you knew it was 27ft at sea level and about 100ft at airliner cruising levels, then 5500m=16000 odd feet will be somewhere in the middle.

[Angleofattack]

Use the following formulas for Altimeter calculations

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 (I assume you can calculate this). This will work for the vast majority of all altimetry questions, except PPL questions use 30ft (not 27).

This will give you the exact answers to questions 1, and 3. I think you mean the indicated altitude at an MSA (true) of 12000ft in Q1.

Question 2 is slightly flawed in that radio altimeters operate only up to 2500ft, and actually read height above the surface directly below rather than altitude. If you put this aside you can determine that the ISA dev is -ve, ie colder using the equation above.


The following formula

Height change per hPa(ft)= 96 x T/P

where T= temperature in Kelvin (C + 273), P = Pressure in hPa


In Q4, T= 252K, ISA in C is -21, as 5500m is approximately 18000ft, P = 500hPa

That gives
Height change per hPa = 96x252/500 = 48.4ft

[AVIATROZ]

Q. you intend to overfly a mountain ridge at an altitude of 15000ft AMsL.The average air temp i s 15degree lover than IsA.the sealevel pressure 1003Hpa.which altimeter indication is needed?

Ans--16170ft...............plz explain hw to solve this problem
thanks

[Kima]

Q. You intend to overfly a mountain ridge at an altitude of 15000ft AMSL.The average air temperature (=ISA Deviation) is 15 degree lower than ISA. The Sealevel Pressure (=QNH)1003Hpa. Which altimeter indication is needed?

15 x 15 x 4 = 900
1013 - 1003 = 10 x 27 = 270

900 + 270 = 1170

15000 + 1170 = 16170

Read Angleofattacks post.. and use the formulae
TA = IA + (4 x IA/1000 x ISA Dev) + (27 x (QNH – Subscale))

[Justwant2learn]

Hello guys,

I am currently studying for my cdn atpl and I came across a MET question I cannot seem to grasp. The concept of port/starboard drift increasing/decreasing true altitude respectively. Now the in the spirit of trying to understand the subject rather than simply memorizing the port drift=increasing true alt and starboard drift=decreasing true altitude, I am trying to attempt to understand this concept. Now where I get mixed up is how do you know whats happening because you could get either port or starboard drift depending on where you sit around the low or the high right? Isn't it a matter of perspective? eg port drift: I could be to the east of a low or west of a high if i was on a west east trip right???

:ugh:

[Yara-ma-yha-who]

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.

[Justwant2learn]

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

[aviatros]

Friends, i am stuck with one of a question.can some plz shed some light in solving this question.thanks for the help
Q.Air at T = +16° C and DP = +4° C is forced from sea level over a 10.000 ft mountain range and descends back to sea level on the other side. If the leeward condensation level is observed to be 8.000 ft, what will be the final temperature?
A)18° B)22° C)20° D)24°
ANS IS B.

[Da-20 monkey]

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:

[SkillsToBurn]

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!

[pudoc]

Length of MAC = 134.5, so:

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

[keith williams]

You need to understand the meanings of the various terms.

The MAC is the Mean Aerodynamic Chord Length.

The statement that the MAC is given as 134.5 inches 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.

[Northbeach]

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

[low n' slow]

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?

[Genghis the Engineer]

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

[Lightning Mate]

As I am a teacher in the subject,

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

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

[paco]

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.

[Genghis the Engineer]

real life has nothing to do with it!

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

[rmcb]

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. :(

[david_320200]

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.

[david_320200]

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

[paco]

The worst place to be is in a cold low.

[KandiFloss]

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 :)

[ChriSat]

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:

[pilotmike]

Your problems started with your very first statement:

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


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

[KandiFloss]

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:

[maximus610]

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! (http://www.flickr.com/photos/76563447@N06/6870955947/in/photostream/)

Thanks!

Mxms

[keith williams]

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?

[schakalix]

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

[Dick Whittingham]

Ola peruano

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

Dick

[schakalix]

ola gringo :)

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

S

[Dick Whittingham]

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

[schakalix]

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

[All-The-Nines]

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.

[keith williams]

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.

[All-The-Nines]

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?

[keith williams]

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.

[Anders S]

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

[keith williams]

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.

[Lightning Mate]

Blimey Keith!

You must be bored. ;)

[clkorm3]

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

[Anders S]

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

[keith williams]

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%

[Anders S]

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

[keith williams]

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.

[Anders S]

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

[keith williams]

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.

[Anders S]

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

[clkorm3]

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?

[keith williams]

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

[clkorm3]

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.

[Anders S]

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

[keith williams]

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.

[Anders S]

Thanks Keith.

-A

[CP Pilot]

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:

[Anders S]

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

[Frank321]

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

[shoeless]

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

[XiRho]

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.

[shoeless]

Yea correct answer is B!!

[mehrabi]

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

[Haroon]

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

[Advs]

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

[Haroon]

can you show please how you are solving it. thanks

[keith williams]

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.

[Haroon]

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

[keith williams]

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

[Haroon]

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:

[keith williams]

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.

[Haroon]

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:

[BGS_Nav]

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!

[Haroon]

Thanks BGS

[Transsonic2000]

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

A Boeing 747-400 has a wing span of 64.4 m. When considering relevant obstacles for the take-off flight path what is the semi-width of the zone at a distance of 1500 m from the TODA?
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

[Haroon]

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)

[Transsonic2000]

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

[manu848]

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:

[bradders147]

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?

[keith williams]

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

[Dick Whittingham]

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

[Anders S]

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?

[keith williams]

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



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?

[Dick Whittingham]

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

[Transsonic2000]

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!

[mad_jock]

Its a design function in the relevant certifying specs

[anthonyd]

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?

[akafrank07]

The general principle of all inertia navigation systems is that the system measures the aircraft's inertia movement from an initial position as a great circle track direction and distance to continuously determine it's up-to-date position.
The compnents of an INS are
1 Accelerometers
2 Gyroscopes
3 Position Computer

The aircraft moves in three dimensions, but the navigation equipment is only interested in acceleration in the horizontal plane. Therefore, the key to the whole INS arrangement is the accelerometers.

With this point;
"The aircraft moves in three dimensions, but the navigation equipment is only interested in acceleration in the horizontal plane. Therefore, the key to the whole INS arrangement is the accelerometers."
How does the aircraft measure if accelerating when climbing etc it is only interested in acceleration in the horizontal plane?

[galaxy flyer]

Quite simply, it does measure acceleration in all three axes, the navigational output to the pilot is in two.

GF

[FlyingStone]

Basically, like galaxy flyer said. INS uses stabilised platform which should be always parallel to earth's surface - so that the accelerometers only measure horizontal movement. IRS is much better, because it uses mathematical calculation to correct output from the accelerometer by using the attitude information - it's easier to do a little calculation than to have a mechanically stabilised platform completely parelel to the earth's surface for the entire time of flight.

[akafrank07]

Thanks guys :)

[akafrank07]

What it is;
"Schuler loop i only common to a stable-platform INS that has been programmed to remain horizontal as the aircraft moves around the surface of the earth. The error exists at the first accelerometer level, i.e., acceleration, which can be passed up through the integration to affect velocity and distance, resulting in a distance error during the schuler loop cycle, but the error returns to zero at the end of the cycle."
Can anyone come up with a simpler explanation of schuler loop, i've tried trusty google but still finding schuler loop hard to understand
Thanks

[Avionker]

http://www.pprune.org/tech-log/119072-ins-irs-schuler-effect.html

Have a read of this if you haven't already, it may help.

[Genghis the Engineer]

The earth wobbles, at a frequency of about 1 wobble per 78 minutes.

An INS doesn't naturally wobble, so therefore it appears to wobble relative to the earth.

This wobble is called the Schuler cycle.

A Schuler loop is a resonant circuit set to the period and amplitude, but in antiphase, to the Schuler cycle.

Adding the output of the Schuler loop to the INS output makes the output correct, relative to the surface of the earth.

G

[Dick Whittingham]

But note that a strapdown IRS keeps a mathematical equivalent of the stable plaform and reacts to setup errors or gravitational errors in the same way as a stable platform INS - with bounded position errors at the Schuler frequency

It is not true to say an IRS does not display Schuler frequency errors

Dick

[Dick Whittingham]

Akafrank07

Whatever referece notes you are using, I suggest you get a better set.

Schuler loop i only common to a stable-platform INS that has been programmed to remain horizontal as the aircraft moves around the surface of the earth

That statement is not true

Dick

[lasseb]

A lot of rubbish has been said about the schuler oscillation... Much of it found in pilot literature...

It is actually very simple.
If you have a platform that is horizontal when you are in London, it will not be horizontal anymore when you reach Australia, if you keep it in a fixed position (relative to space).
And gyros provide fixed positions relative to space only. So we need some sort of mechanism that will tie the platform relative to the round earth.

In INS systems this is done by measuring acceleration and thereby the speed that we move with.
So if you are moving with XX kts direct south, a tie-mechanism will tilt the platform xx deg/hour so it is kept horizontal with regards to the earth.

So far so good... The problem is then, that this tie mechanism has no clue weather an acceleration is actually caused by the airplane moving, or because the platform is not horizontal.
If the airplane is parked, and the platform for some reason is not horizontal, you will pick up an acceleration due to earth gravity.. This will be interpreted as if the plane was actually moving, and the tie-mechanism will start tilting the platform. The tilting will continue until the platform passes the horizontal, and starts tilting the other way. This is then interpreted as a deceleration, and eventually an acceleration in the opposite direction, causing the tie-mechanism to now tilt the platform back again...And we then have an oscillation....
The frequency of this is quite slow (84 minutes), and unless you dampen this the INS/IRS wold be worthless.
Schuler was a German engineer who claimed that INS/IRS system would be impossible to build, because an accelerometer, is unable to figure out if it is measuring actual acceleration or "false" acceleration due to a tilted platform. He actually published a science paper about this :-D

[Dick Whittingham]

lasseb

If you "dampen" the response to errors due to incorrect levelling how do you allow the primary function to continue un-damped?

[lasseb]

You are not dampening a response, and it is not incorrect leveling. The platform will always oscillate. That's the nature of a feedback circuit. The idea is just to subtract a known (error) oscillation from the incoming signal.

What you are doing is deducting a very low amplitude/low frequency signal from the input. This signal is (or rather should) correspond to the oscillation of the platform. The signal is generated internally in the INS/IRS computer, based on a math-model of the system.

Measuring raw input from the platform when an aircraft is parked would tell you that the aircraft is moving forth and back (both N/S and E/W) with an 84 minutes oscillation period. Basically it would tell you that you where driving around in a circle ;-). The INS/IRS computer has a model of this signal internally and deducts it from the input signal. If the modeled signal is a perfect match to the actual oscillation of the platform, you'll have a perfect INS/IRS, that tells you that you are actually parked when you are parked ;-)

This compensating circuit is working both in platform and strap-down units. It's just easier to imagine using a platform. but the accelerometer errors exists in both scenarios.

To add confusion, some models of the INS actually fed the compensating signal to the platform leveling system in stead of deducting it from the input. This will cause the platform to hold still even in parked scenarious. I guess that computers where to slow these days to handle all the calculations, so the more that where done mechanically the better.
Both approaches will work.

EDIT:
The best explanation out there I have found is a NASA document describing low orbiting satellites. They have an initial paragraph on schuler. Very good stuff..

I am apparently to stupid to figure out the link to the PDF, but search for "SCHULER PERIOD IN LEO SATELLITES ", and you will get it.

[Dick Whittingham]

lasseb

Continuing, for simplicity, with the stable platform INS, if in ALIGN the levelling is carried out without error and there is no gravity element being sensed by the N/S and E/W accelerometers, why would the platform oscillate?

I think you are saying that if there is a leveling error and the platform is oscillating then the response is memorised, carried forward into the NAV regime and removed from all subsequent navigation computations. Do I read you correctly?

Edit. I think I see that you are assuming a rough initial alignment leading to error which is then computed out or, in older systems, fed back to the platform control. Am I getting warmer?

[lasseb]

Hi Dick
This is actually harder to describe en words than I thought :-D.

What I'm trying to emphasize here is that the schuler error signal is not read from the platform, or memorized, or anything, it is strictly calculated/generated in the computer (based on a math model) and deducted from the incoming signals from the accelerometers almost at the very start of the calculations.

So the signal going to the integrators is = (SENSOR_DATA - SCHULER_DATA_MODEL.)

Regarding initial oscillation you are kind of correct. In theory if the platform is perfectly aligned no error should ever be present, and the platform should never oscillate. There are 2 problems with this. The first is that there is no such thing as perfect, and even small noise levels in the signal wiring will introduce this error. Also mathematical noise in the computer will do this, since we are not dealing with infinite decimals.

Secondly, because we have introduced the schuler compensations, that will actually make the platform oscillate if starts with no oscillation :-D.

Consider the following:
Lets say that the platform is perfectly aligned, dead-level, perfect wires, perfect computer, no oscillation.
The signal input to the integrators, are now the signal from the accelerometers - (which is zero), but deducted for the current schuler period value. So if the platform is perfect level, the input to the integrators would give us the 84 min oscillation signal (or rather the inverted value). This would then lead to the integrators giving us speed, which leads to the tie-mechanism trying to tilt the platform, and then we have the oscillation...

You could say, that because we have introduced the schuler compensation, the platform must oscillate all the time,
If we where only flying around in a 200NM radius (or so) from home base all the time, we could skip the entire schuler compensation mechanism, and use the same tie-mechanism as an attitude indicator. We would then only need to compensate for heading, and it would be much easier.

Hope this answers the question :-)

[Dick Whittingham]

Lars

Yes, it's a word thing. I now think I understand what you say, and I am happy to find that I have not been wildly wrong, Thanks a lot

[akafrank07]

Thank a lot of the explanations - genesis the engineer Dick and Lasseb, i now understand it a bit better now, but obviously a quite complicated subject! Dick and Lasseb you two guys know your ****.:)
N.B. Dick i got this from 'ace the tech pilot interview' i understand there is meant to be a few mistake from this book, but sadly it seems to be what the airlines are using, as a couple of my friends that are now with the airline studied this book religiously though for my own sake and others reading trends i am trying to find out where these mistake are in the book, thanks for your help Dick.

[Anders S]

Hey guys,

What is the vertical separation minimum below 30,000 ft?
a. 500 ft.
b. 1,000 ft.
c. 2,000 ft.
d. It depends whether or not RVSM is applied.

The correct answer is; b. 1,000 ft, and the reference states:

The vertical separation minimum (VSM) is:
• Within designated airspace (subject to RAN agreement (RVSM)), a nominal 300m (1,000 ft) below FL 410 or a higher level where so prescribed for use under specified conditions, and a nominal 600m (2,000 ft) at or above this level and; and
• Within all other airspace: a nominal 300m (1,000 ft) below FL 290 and a nominal 600m (2,000 ft) at or above this level.

This however doesn’t appear to coincide with the flight levels stated in the ICAO tables (the semi circular rule), in which case you can have an east bound VFR flight on FL 35, a west bound VFR flight on FL 45 and a west bound IFR flight on FL 40, giving a vertical separation of 500 ft.

Am I missing something here?

[hvogt]

Anders,

what you are missing is that VFR flights are only seperated in airspace classes A through C (cf. ICAO Annex 10, Appendix 4). Apart from this, you will seldomly find IFR traffic at FL 40 in uncontrolled airspace. Firstly, FL 40 would only exist in a hypothetical country having a transition altitude which is low enough, and secondly such a low flight level would mostly be in uncontrolled airspace.

Edit: Sorry, the part after 'secondly' is rubbish. Forget it, please.

[Anders S]

Ok, so basically it’s not considered “separation” outside of controlled airspace?

Those flight levels were just examples, the principle would still stand, I think, if flight levels 135, 145 and 140 were used instead.

Thanks for prompt response.

[Anders S]

Hey guys, I have a question for you.

Look at this TAF for Zurich airport
TAF LSZH 211322 22018G35KT 9999 SCT012 BKN030 BECMG 1315 25025G45KT TEMPO 1720 4000 +SHRA BKN025TCU BECMG 2022 25015KT T1815Z T1618Z=

Which of these statements best describes the weather likely to be experienced at 1500 UTC?

B. Meteorological visibility 10 km or more, main cloud base 3000 ft, wind 250°, temperature 18°C

C. Meteorological visibility 10 km or more, main cloud base 1200 ft, gusts up to 45 knots.

The correct answer is supposedly B, whereas C was my selection.

There is the matter of the entries “T1815Z T1618Z” at the end of the TAF which I’m not entirely sure what they mean, but the book states that “TAFs do not contain information on temperature”, and I also ran through the CBT on the subject and couldn’t find anything about it there either, so am I missing something?

[Da-20 monkey]

Hi Anders,

I edited my previous entry about ,,main cloud base'' being the same as cloud ceiling:

Ceiling is by definition more than SCT, but it turns out that main cloud base is the higher layer if there are multiple layers, and the higher one is thicker.

I pm you a thread from another forum about this very question.

[hvogt]

Anders,

my interpretation of the respective group is the forecast temperature for 1500H UTC is 18 °C and the forecast temperature for 1800H UTC is 16 °C.

ICAO Annex 3 states forecast temperatures, though not an element of a standard TAF, may be included in accordance with regional air navigation agreements. If they are given in a TAF it is recommended to indicate the maximum and minimum temperatures for the forecast period.

The format of the temperature group in your example, however, is not in line with ICAO Annex 3 nor the WMO Manual on Codes in their current versions. I take it the example is from a question bank and might be outdated.

If the temperatures in your example were indeed maximum and minimum temperatures at 1500H and 1800H this would have to be indicated today as "TX18/1500Z TN16/1800Z", with "TX" standing for maximum and "TN" standing for minimum temperature.

[Anders S]

Ok, that makes sense. Thanks for the responses.

[PTCstudent123]

Hi

there are a few questions that are confusing me right now with regards to gen nav.

these are departure and convergency. I never seem know which formula to use and when. I understand the 2 formulas. Ch long x sine mean lat and ch long cosine lat. can someone give an example of when I should use both of these.

Right now I'm looking at q' 5566 in the atpl online system.
Given: Position 'A' N60 W020, Position 'B' N60 W021, Position 'C' N59 W020. What are, respectively, the distances from A to B and from A to C?

Why do I use the departure one for this question and not the convergency?

This is probably a pretty stupid question but its totally confusing me.

Thanks in advance

[Da-20 monkey]

Hi,

Use the departure formula for your example, because it is for calculating distance between two point on the same latitude.

You use the convergency formula for calculating change of direction of a great circle track. Not for calculating distance, unless perhaps the average latitude and the convergency is given, in which case you could calculate the change of longitude.

To put it simple, departure and convergency has little to do with each other. One is for distance, the other for degrees.

[RichardH]

In simple terms if the question is mentioning the words distance or position it is highly likely to be a DEPARTURE related question, going east/west along a parallel of latitude.

So in your example,which funnily enough I worked through with a student yesterday. A to B is a departure question. So Dep = Ch Long (mins) x cos lat.
1 * 60 = 60 * cos 60 (.5) = 30 nm. However A to C is not departure as the positions are on the same meridian. It is testing do you know at 1 degree of Latitude is 60 nm so A to C = 60 nm.

Now if the question mentions the words like great circle, rhumb line, initial or final track highly likely to be a CONVERGENCY problem. I always teach my students tips similar to the ones mentioned.

[akafrank07]

TOTAL PRESSURE ERROR

"As an aircraft moves through the air, a static pressure disturbance is generated in the air, producing a static pressure field around the aircraft. At subsonic speeds, the flow perturbations due to the aircraft static pressure field are nearly isentropic and do not affect the total pressure. As long as the total pressure source is not located behind a propeller, in the wing wake, in a boundary layer, or in a region of localized supersonic flow, the pressure errors due to the position of the total pressure source are usually negligible. Normally, the total pressure source can be located to avoid total pressure error."

"As an aircraft moves through the air, a static pressure disturbance is generated in the air, producing a static pressure field around the aircraft."

What is this static pressure field around the aircraft i would of thought this was just the free air around the aircraft, the air which is not impacted by the aircraft which would be then dynamic pressure right?

It then goes on to say: "at subsonic speeds, the flow perturbations due to the aircraft static pressure field are nearly isentropic and do not affect the total pressure."

I understand that isentropic basically means the same all round though i can't grasp what this paragraph mean either can anyone simplify it?

[RTN11]

Basically there is a thin layer around the aircraft in which it feels like static pressure - a good way to demonstrate is put your hand out of the window in a moving car and put it flat against the door, you will feel no dynamic pressure, only static. Hence, static ports are usually small holes on the side of the fuselage.

To measure the PITOT or toal pressure, the Pitot tube must stick out far enough so it it outside this static layer. Remember that PITOT (total) pressure is dynamic pressure PLUS static. For altitude read out you need static pressure, for airspeed you need dynamic, dynamic pressure is created by the movement through the air.

[Jetpipe.]

"As an aircraft moves through the air, a static pressure disturbance is generated in the air, producing a static pressure field around the aircraft."

Some thoughts..
As you probably are familiar with the bernoulli effect (Pt=Ps+Pd), the distribution of dynamic and static pressure over a moving surface depends on many variables (aoa, airflow angle, camber, material, etc.) f.ex. at the stagnation point (http://en.wikipedia.org/wiki/Stagnation_point) of a wing the dynamic pressure will be 0 and the static equal to Total! So the a/c form can interfere with the ''normal'' airflow and that will cause pressure disturbances, mostly increased static pressure because of the deceleration of the air molecules that come in contact with the airframe . At subsonic speeds the air is thought to be isentropic, meaning that it can adiabatically change state (Kinetic to Potential energy without heating loss and vice versa). At supersonic speeds and above (I think even transonic), the air can not longer be seen as an isentropic, incompressible fluid and that means it will produce an error that will affect Total pressure readings!

Hope I didn't mess it up! ;)

[PTCstudent123]

Thanks Richard and Da-20 monkey that explains things better. as you can see by my user name I am an ex PTC student and I am now doing atpl's on my own for the time being. It is hard trying to figure out everything on your own :)

thanks again

[selfin]

... PITOT (total) pressure is dynamic pressure PLUS static. [F]or airspeed you need dynamic [] pressure [] created by the movement through the air.

Total pressure in a compressible medium is not dynamic + static pressures.

When air (M ≤ 1) is brought to rest isentropically at the head of the pitot probe, the total pressure will be given by a function such as (6) in the graphic on this Glenn Research Center page, Isentropic Flow Equations (http://www.grc.nasa.gov/WWW/k-12/airplane/isentrop.html)

The subject of (6) in the above is Ps/Pt (s=static; t=total) and will always take a value equal to or less than unity. Hence, Pt ≥ Ps or Pt - Ps ≥ 0. This latter quantity, the difference between total and static pressure, is often said to be dynamic pressure. It is in fact something else called impact pressure. If you look at (5) in the above reference, which is the definition of dynamic pressure (defined in this form by Euler and not Bernoulli), you can see the difference in the behaviour of these two functions.

Airspeed indicators map impact pressure (not dynamic pressure) to calibrated airspeed. The function which provides that mapping assumes an isentropic process, meaning as given by Jetpipe that the process leading to a total pressure is an adiabatic and reversible one, i.e. heat is neither added nor removed and no dissipative effects occur. Obviously if these conditions are not satisfied the airspeed value will be erroneous. For these reasons it is quite important that both the pitot probes and static ports are designed, and located, to ensure minimum dissipation and heat addition.

[akafrank07]

Hi Jetpipe



"As you probably are familiar with the bernoulli effect (Pt=Ps+Pd), the distribution of dynamic and static pressure over a moving surface depends on many variables (aoa, airflow angle, camber, material, etc.) f.ex. at the stagnation point (http://en.wikipedia.org/wiki/Stagnation_point) of a wing the dynamic pressure will be 0 and the static equal to Total! So the a/c form can interfere with the ''normal'' airflow and that will cause pressure disturbances, mostly increased static pressure because of the deceleration of the air molecules that come in contact with the airframe . At subsonic speeds the air is thought to be isentropic, meaning that it can adiabatically change state (Kinetic to Potential energy without heating loss and vice versa). At supersonic speeds and above (I think even transonic), the air can not longer be seen as an isentropic, incompressible fluid and that means it will produce an error that will affect Total pressure readings!"

Why can the air no longer be seen as an incompressible fluid at supersonic speeds and how come it starts to lose heat at supersonic speed?

[Jetpipe.]

The air IS a compressible fluid but at low speeds the friction heating is almost negligible so the isentropic equations for incompressible fluids are quite so precise... Check this wiki-page about Total air temperature (http://en.wikipedia.org/wiki/Total_air_temperature) !

If you need a more academic answer, I think Selfin is the man to ask! :ok:

[akafrank07]

PITOT (total) pressure is dynamic pressure PLUS static. [F]or airspeed you need dynamic [] pressure [] created by the movement through the air. Total pressure in a compressible medium is not dynamic + static pressures.

When air (M ≤ 1) is brought to rest isentropically at the head of the pitot probe, the total pressure will be given by a function such as (6) in the graphic on this Glenn Research Center page, Isentropic Flow Equations

The subject of (6) in the above is Ps/Pt (s=static; t=total) and will always take a value equal to or less than unity. Hence, Pt ≥ Ps or Pt - Ps ≥ 0. This latter quantity, the difference between total and static pressure, is often said to be dynamic pressure. It is in fact something else called impact pressure. If you look at (5) in the above reference, which is the definition of dynamic pressure (defined in this form by Euler and not Bernoulli), you can see the difference in the behaviour of these two functions.

Airspeed indicators map impact pressure (not dynamic pressure) to calibrated airspeed. The function which provides that mapping assumes an isentropic process, meaning as given by Jetpipe that the process leading to a total pressure is an adiabatic and reversible one, i.e. heat is neither added nor removed and no dissipative effects occur. Obviously if these conditions are not satisfied the airspeed value will be erroneous. For these reasons it is quite important that both the pitot probes and static ports are designed, and located, to ensure minimum dissipation and heat addition.

When air (M ≤ 1) is brought to rest isentropically at the head of the pitot probe

What does isotropically mean?


For these reasons it is quite important that both the pitot probes and static ports are designed, and located, to ensure minimum dissipation and heat addition

Does the location affect heat addition because the pitot probe is located far out from the aircraft skin and the static probe is built into the aircraft?


Cheers

[Anders S]

Hey guys, need some help again.

Question 1:
On a CVOR the phase difference between the AM and FM signals is 30°. The VOR radial is?
a. 210
b. 030
c. 330
d. 150

Question 2:
For a conventional VOR a phase difference of 090° would be achieved by flying …… from the beacon.
a. West
b. North
c. East
d. South

I put down b. on Q1, which was incorrect, and c. on Q2 which was correct. According to the answers the correct answer for Q1 was c. 330. The only thing I can come up with is that it’s in the “wording” of the question, because they are asking for the difference between the AM and FM signals, and not the difference between the FM and AM signals. Could it be that simple, or am I missing something else?

[keith williams]

Did question one include the words TO or FROM?

The 30 degree phase difference will occur on the 030 radial FROM the VOR, which is also the 210 radial TO the VOR.

[Anders S]

Nope, the question is basically copy/pasted.

[keith williams]

In my copy of CQB15 the correct answer is marked at 030. Is it perhaps just a case of your reference material being wrong?

[Anders S]

Quite possibly, it wouldn't be the first time.

[PTCstudent123]

Hi can someone please explain this question to me. I don't really understand how it isn't a track change of zero. Its following the same line of latitude but the longitude is changing. wouldn't this mean that its just following a straight line towards the east?

The following points are entered into an inertial navigation system (INS). WPT 1: 60°N 30°W WPT 2: 60°N 20°W WPT 3: 60°N 10°W The inertial navigation system is connected to the automatic pilot on route (1-2-3). The track change when passing WPT 2 will be approximately:

4° decrease
9° increase
zero
9° decrease

Some help here would be great thanks






Am I right in thinking that I need to find the conversion angle?
I.E. find the convergency and half it? that way I know that for every 5degrees it will decrease by that number?

[Skittles]

In regards to the question about the CVOR and 030 - the difference between AM and FM is 30 degrees, but I think I'm right in saying that the phase difference to be measured is between FM and AM, not AM to FM.

AM to FM is 'backwards,' and given that the phase scanning is clockwise in a CVOR you'd have to go right round - i.e. 330 degrees.

In relation to the Gen Nav question, you are following great circle tracks. I assume you've learnt about earth convergence (sin mean lat) and conversion angle (1/2 sin mean lat).

You're in the northern hemisphere so your initial track to wpt 2 will be less than 90, and the final track more than 90. At wpt 2 your heading will decrease again to maintain another great circle. I imagine the answer will be D!

[PTCstudent123]

thanks Skittles This is what I was thinking but I couldn't figure it out at first. I came across a similar question Which had a better worded explanation.

Thanks mate

[Skymaster15L]

Hello everyone,

The efficiency of a gas turbine engine increases with:
A) an increase in ambient temperature
B) an increase in volumetric efficiency
C)a decrease in ambient air pressure
D) a decrease in ambient air temperature

While A &C are obviously wrong, having already gotten this question incorrect, could you please tell me why the answer more fitting is D and not B? I understand that a decrease in air temp increases air density, which has the effect of "naturally" increasing mass flow thorough the engine, without making the engine work more to artificially produce (faster compressor spin rate) the same amount of thrust for any give throttle position. Mass flow and acceleration imparted upon a given mass of air are after all the main factors which influence thrust production as I have been taught.

Now, isn't "an increase in volumetric efficiency" a better choice and an overall better answer in terms of terminology since it implies than an engine with better VE has better mass flow through the engine at all times, in any given ambient conditions compared with one of less VE? Thus, the answer can cover a broader range of all variables than just lower amb. temp, although this answer is technically correct as well.

Or is this term more generally reserved to performance discussion relating to piston engines and should not be used with gas turbines?

Thank you

[Kiss My APU]

Volume doesn't tell you the mass. For that you'd need to know more about its density. With very hot air you may have very little mass and vice versa with cold air. This is to do with the link between temp. and density. Therefore, a colder temperature will give you a higher density which is a greater mass per volume and therefore greater thrust.

A measurement of volume alone isn't enough information.

*I'm only studying myself so hopefully others will contribute and confirm my suspicions.

[Dick Whittingham]

First, differentiate between efficiency and thrust. You can have a grossly inefficient engine producing lots of thrust. Second, efficiency is always stated as a comparison - SFC is a good example, fuel flow against thrust produced.

So the question starts of badly. What is "the efficency" of a jet engine? We could look at thermal efficiency, volumetric efficiency or propulsive efficiency, to name but a few. But in this context - ATPL questions - it probably means SFC

So how do the answers stack up? Decreased intake temp does two things. It increases density and therefore mass flow and thrust but also increases thermal efficiency. Themal efficiency is driven by the difference between the fixed EGT and the inlet temp. Volumetric efficiency is driven by engine design and is usually a maximum at about 90%N1. It is, therefore under your control to some degree but I don't think the examiner means that. At the back of the examiner's mind he is linking this with the conditions for max range, when you would use 90%N1 as a given. That leaves increased pressure, which would increse density and therefore mass flow and thrust but will have no direct efect on "efficiency"

A bit woolly, I'm afraid, but I hope it helps.

[Skymaster15L]

Better a wooly answer than a scantily clad one sometimes, in this case it does actually make sense. The trickiest thing I've found about some questions (also valid when examining graphs) is not necessarily the question itself that is being asked, but the associated assumptions/conditions the question is set against, that the question doesn't tell you about and can potentially affect a correct answer being given.

Thanks guys

[clkorm3]

A small (1ft x 1ft6in x 2ft 3in) heavy crate, mass 500lb is to be transported in mep specimen aircraft zones 2 and 3. Based on the maximum floor loading limit of 120lb/ftsquared can a box be loaded directly into the aircraft?

[Da-20 monkey]

max floor loading limit is 120 lbs/ft². Area of crate is 1ft x 1,5ft= 1,5ft².

So:

120 lbs ...180 lb/ft²
_____ = _______

1 ft².........1,5 ft²

The 500lbs crate is heavier than the maximum for the 1,5ft² area, so you cannot load it.

However, I heard a retired redcap say once that in real life they multiply the cargo masses with 3 to get the actual cargo weight at the max load factor of 2,5 for transport aircraft+ 0,5g extra margin.

[clkorm3]

Thank you man

[PTCstudent123]

question 3286 (atpl online)

Hi

I am trying to figure out this question and I dont understand the explanation of the question. can someone help me please.

Rotate to put track 090ºT next to HEADING index; read drift (12º right (Stbd)) and rotate to put track (090º) under drift (12ºStbd).
Drift has now changed to 13ºStbd so put track 090º under 13ºStbd.

Why do you have to rotate it again( in blue) What does this do? When I do it I get 11 degrees.

Thanks

[Skittles]

You're doing a drift shuffle.

If you look at the tope of your CRP-5 you'll notice that it says 'true heading' not 'true track.' When you first put the 90 under the true heading that's wrong, you've lined up the track with the heading marker. This doesn't take into account any drift.

To resolve this you have to do a drift shuffle, which is the process described in your explanation. Check the drift on the centre bit and then rotate the wheel in the same direction until the true track is on the relevant number. In your example you had 12 starboard drift, so you have to put 90 under 12. Then you have to check the drift again. Has it changed? In your case it had - it had changed to 13. Do the same process, move the 90 under 13. You keep 'shuffling' like this until it all lines up (i.e. the 90 is under the 13, and the drift shows as 13), then the value under the 'true heading' is the heading you have to fly to make good that track.

Remarkably difficult to explain via text, but if you google the CRP-5 drift shuffle you'll get better explanations.

[RichardH]

It would be useful to copy full question and answers into forum as not everybody is on ATPL online.

That said "Skittles" has explained it. However I should like to add a couple of pointers.

The correct term is "balancing drift" and it only needs to be done when working out a heading and/or ground speed when using the wind down method. Do NOT balance/shuffle for any other problem.

As the correct term suggests you have given a track/course you have to turn into wind to allow for it. It is this very turn that changes the original angle of the wind which may lead to having to "shuffle" a couple of times. You ALWAYS turn in the direction of the dots new position.

You are "balanced" when your dot on the drift lines = the number of degrees (heading to track) difference in the opposite direction.

There is another & simpler way if you care to PM me.

[Skymaster15L]

"A stagnation point of a body in a moving airstream is a point where: (answer) the velocity of the relative airflow is zero and the surface pressure is higher than the ambient atmospheric pressure"

Could you please tell me why surface pressure is higher and not equal to ambient atmospheric pressure? As the book that I read on this says, (perhaps also failing to expand more on this topic),that for incompressible flows, the total pressure of still air is composed of 100% static pressure with 0 % dynamic pressure if there is no movement/velocity involved. As soon as there is velocity given to a body of air, static pressure starts to drop, and total pressure is now also composed of dynamic pressure. The sum of these pressures, P total, is always a constant.

Also, since at the stagnation point where local airflow flow velocity is 0, I take it that static pressure is then highest, meaning no dynamic pressure is present, so shouldn't this pressure then be equal to ambient air pressure of the nearby parcel of air that is unaffected by the wing's passage through it?

I case that I am possibly:
1)using the terms static pressure and stagnation pressure interchangeably, when they actually mean different things, or
2)worrying about a question that was written wrongly to begin with, or
3)not understanding something else that I have failed to mention or learn

How can the surface pressure at the stagnation point be higher than ambient, when the sums of P Static and P Dynamic, in whatever proportions, are always a constant anyway. What is adding that extra pressure?

Appreciate your input

[packo1848]

Anyone feel free to correct me if I'm wrong here anyone, I've just done the same lesson, and I understand the stagnation point to be the point at which the air is brought to rest on the aerofoil. Therefore it would be total pressure i.e. Static (or ambient) + Dynamic, the reason being the Kinetic energy has to be transferred from the airflow to the aerofoil in order to bring it to rest.

[keith williams]

Total Pressure = Static Pressure + Dynamic Pressure.

We need to be careful in using the statement that "Total pressure is constant".

It is more correct to say that: "Total pressure is constant at all points in a stream tube provided no energy is added to or subtracted from the airstream."

To understand this let's look at an aircraft sitting on the runway in still air at ISA mean sea level. Static pressure is approximately 15 PSI and because the air is still, the dynamic pressure is zero. So the total pressure is approximately 15 PSI.

If we now accelerate the aircraft, the dynamic pressure will increase with the square of the TAS. Eventually a speed will be reached at which the dynamic pressure is 15 PSI. But this does not mean that the static pressure has fallen to zero. The increased dynamic pressure was caused by the extra energy that we provided by accelerating the aircraft. So the ambient static pressure will still be approximately 15 PSI and the total pressure will be approximately 30 PSI.

If we continue to accelerate the dynamic pressure will become greater than 15 PSI, so the total pressure must be greater than its initial value of approximately 15 PSI.

At the stagnation point the airflow is brought to rest and this converts all of its dynamic pressure into static pressure. The total static pressure (the stagnation pressure) at the stagnation point is then the sum of ambient static pressure plus the dynamic pressure that has been converted. In the case of our aircraft flying at a speed at which dynamic pressure is 15 PSI we would have 30 PSI of stagnation pressure, but only 15 PSI of ambient static pressure.

Later on in your studies you will look at the effects of shock waves. When air flows through a shock wave it is abruptly compressed. This converts some of the pressure energy into heat. This reduces the total pressure. So as air flows through a shock wave the total pressure of the air stream decreases.

[Skymaster15L]

the part that "provided no energy is added or subtracted from the airstream" does shed light on the light on the topic now, and I wish that was included in the book to begin with. Just a few more questions about this however:

Does this mean that the Total P=Static P+Dynamic P formula now mean that the T. pressure being a constant only really applies to an air mass with Static or Dynamic pressure fluctuations only caused by the wind/ breeze blowing? As soon as an airplane passes thru such an air mass, it blows everything out of the water in terms of validity/simplicity?

Also, what happens to the value of total pressure in other areas of the wing, for example the low pressure region responsible for lift? Does this value also exceed ambient pressure because we are adding energy to the airflow, or does airflow here behave more like in the above said equation since the airflow velocity does not reach zero like at the LE stagnation point?

My question being rephrased to ask that: does the Total Pr. of 30 PSI in our hypothetical example then, reach such a value higher than ambient, only in the region of the stagnation point because of the unique nature of its airflow, whereas at any other points along the wings, the variable constituents of static and dynamic pressure are always a constant (and equal to ambient pressure) but never higher?
Thanks for the clarification

[keith williams]

This whole subject is related to the principle of conservation of energy. If we assume that no energy is added to or removed from our air stream, then the total energy must remain constant.

If we look at a moving mass of air we can see that it's total energy is the sum of the following parts.

1. Mechanical energy due to its static pressure. (this is the energy that permits compressed gasses to carry out work when they expand, as for example in a piston engine).

2. Kinetic energy due to its velocity (this is evident in the form of the dynamic pressure).

3. Potential energy by virtue of its height above some reference point.

4. Thermal energy (this is evident by virtue of its temperature).

If we simplify our experiment by assuming that the flow is horizontal, then the potential energy will be constant.

If we now introduce our air stream into a tube at some selected velocity, its total pressure energy will be the sum of the static pressure plus the dynamic pressure.

If the tube becomes narrower, the velocity must increase to permit the flow to continue at the same mass flow rate. This will increase the dynamic pressure.

But(if we ignore friction) the narrowing of the tube will not introduce or remove any energy, so the total pressure energy must remain constant. This means that the static pressure must decrease to offset the increasing dynamic pressure. The above scenario is an example of a situation in which the total pressure remains constant.

Now let's look at what happens when an aeroplane accelerates during the take-off run. Its engines provide a great deal of energy to accelerate it up to flight speed. This increases the dynamic pressure. But this increase in dynamic pressure is being funded by the engines, so there is no compensating reduction in static pressure. The overall result is that as the aircraft accelerates, the total pressure of the air flowing over its surfaces increases.

But if we look closely at its curved surfaces we can still see that localized trade-offs between static pressure and dynamic pressure are taking place. As the air flows over the nose for example, the curved surfaces cause the local airspeed to increase. This increases the local dynamic pressure and decreases the local static pressure. But these local changes are being applied to a total pressure which is already much higher than the ambient value.

[Skymaster15L]

Thanks for setting me straight on this K.W. It's nice to finally bridge a gap btwn what textbooks present to you in simplified format, and what is actually going on. I am revising and expanding on previous knowledge not only for the ATPL exams I am studying for, but also to slowly set aside a folder of detailed notes & explanations on the most important knowledge topics for when I start my FI rating.

Cheers :ok:

[Anders S]

Hey guys.

I don’t get this one, and I have seen several explanations for it, among others right here, but in another thread, but I haven’t considered them to be satisfactory, so I’ll give it another go.

Given: Waypoint 1 is 60°S 030°W, waypoint 2 is 60°S 020°W. What will be the approximate latitude on the display of an inertial navigation system at longitude 025°W?

a. 60°06’S
b. 59°49’S
c. 60°00’S
d. 60°11’S

The correct answer is supposed to be, a. 60°06’S.

My solution was:

0.5 x 10° x sin(60) = 4.33°, conversion angle.
5° x cos(60) x 60 =150 nm, departure.

Then do some trig-stuff, or 1-60 rule, whichever, to come up with roughly 11 nm, which would make d. an appropriate answer in my opinion.

Most of the explanations I have seen for this question seem intent on halving the conversion angle, or rather getting the conversion angle from the ch.long between 030°W and 025°W, to get 2.2°-ish, and obviously that would present an answer more consistent with the correct one. But why is this done? I don’t see how that’s relevant as it would imply that we are flying from 60°N 030°W to 60°N 025°W and then on to 60°N 020°W, in which case I would fully agree with the answer, but that would mean that 025°W was an additional waypoint… Unless I'm missing something else.

-Anders

[Lightning Mate]

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.

If you teach the subject then why do you not know the answer!

The water drain is sealed, otherwise there would be a permanent loss of pressure.

[Da-20 monkey]

0.5 x 10° x sin(60) = 4.33°, conversion angle.

You should use 5 degrees here instead of 10. It's obvious if you draw a picture of the situation with 2 parallel lines representing the mentioned lines of longitudes, like on a mercator chart, with straight rhumb line between the points and a downwards pointing great circle line. The point where the g.c. is most south is in the middle between 20W and 30W, so 5 degrees longitude.

Even then I am not sure how accurate this answer is in the real world..

Back in the days we didn''t have this q. in the database. (pipe-smoking smiley)

[Anders S]

Nope, still don’t get it. The way I see it, that would give me the most southerly point between 30W and 25W, ie 27.5W.


Strike that, I'm moling it over and hopefully it will come to me.

-Anders

[Da-20 monkey]

0.5 x 5° x sin(60) = 2,17°, conversion angle.

I mean 5 deg. for change of longitude.

Tangens = opposite/ adjecent

Tan 2,16= opposite/ 150NM

opposite= 5,67NM.

this is appox. 6 minutes of logitude, so 60,06

[Anders S]

I’m sorry, I thought I had it but now I’m just confused again. The way I’ve learnt it is:

Conversion angle = 0.5 x ch.long x sin (lat)

But what you’re saying is basically:

Conversion angle = 0.25 x ch.long x sin (lat)

The change of longitude is 10°, not 5°, as I see it…

-Anders

[bayblade]

anders, i agree with your method and answer. i guess the answer printed is wrong.
the formula for conversion angle is correct as 0.5*ch long*sin(lat)

[bayblade]

i guess i've figured it out:) with some help from another forum.
lets name the starting position A and the intermediate position on 25 degree W as B.
when we join points A and B to form a triangle for calculation purpose, it divides conversion angle into two equal parts of 2.16 degress each.
now solving the sum in the same manner:
tan 2.16= X (unknown)/150
X=5.65nm
X/60 = 0 degrees 5 minutes
hence position B= 60 degrees + 5 minutes
=60 degrees 5 minutes
:ok:

[Da-20 monkey]

like bayblade said

[keith williams]

The midway point between 30W and 20W is 25W.

At 60S the departure between 30W and 25W is 150 nm.

We can now draw a triangle with a horizontal side 60S 30W to 60S 25W. The length of this line will be 150 nm.

The second side is a vertical line from 60S 25W down to the unknown position south of 25W.

The third line is the hypotenuse form 60S 30W to the unknown position south of 25W.

The internal angle between the hypotenuse and the horizontal side is the conversion angle based on going from 60S 30W to 25W.

This angle = 0.5 x 5 degrees x Sin 60 = 2.165 degrees.

The tangent of this angle is the vertical side divided by the horizontal side.

Tan 2.165 = Ch lat / 150 nm

Rearranging this gives Ch lat = Tan 2.165 x 150 = 5.67 degrees.

This means that the change of latitude between 60S 30W and the unknown position south of 25W is 5.67 degrees. This makes the new latitude 60 degrees 5.67 minutes south. The closest option to this is 60°06’S

[keith williams]

Sorry Anders my previous post did not address this part of your question.

Most of the explanations I have seen for this question seem intent on halving the conversion angle, or rather getting the conversion angle from the ch.long between 030°W and 025°W, to get 2.2°-ish, and obviously that would present an answer more consistent with the correct one. But why is this done? I don’t see how that’s relevant as it would imply that we are flying from 60°N 030°W to 60°N 025°W and then on to 60°N 020°W, in which case I would fully agree with the answer, but that would mean that 025°W was an additional waypoint… Unless I'm missing something else.


To understand why they do this we need to sketch the whole picture. Draw a horizontal straight line to represent the 300 nm rhumb line track from 60S 30W to 60S 20W.

Now draw a shallow arc looping down between the two ends of the rhumb line track. This represents the great circle track.

Midway between the two ends of the rhumb line draw a vertical line down to the great circle arc. This vertical line is 150 nm from each end of the rhumb line track. The length of this vertical line represents the maximum change of latitude between the two tracks.

Now draw a sloping straight line from each end of the rhumb line to the lower end of the vertical line.

Using the conversion angle equation we can now calculate the angle between each end of the great circle arc and the ends of the rhumb line.
This is 0.5 x 10 degrees x Sin 60S = 4.33 degrees. Write 4.33 in each of these angles.

Now let’s look at what happens when we fly from 60S 30W to the midway position south of 25W. We are initially tracking 094.33 degrees, but our track is continuously turning to the north, such that we are tracking 090 when we reach our most southerly point.

During this first half of the trip we have reduced our track direction by 4.33 degrees from 094.33 to 090. This means that our mean track was 092.165 degree. This mean track is represented by the straight line from our starting point to our most southerly point. So the internal angles in the triangles at each end of our track are 2.165 degrees. This is half of the conversion angle.

So in solving this type of problem it is easier to go straight for ½ the conversion angle and work out the solution from there.

[Anders S]

I figured it was something like that, but I was having a hard time visualizing it. Thanks for clearing that up, and for everyone else efforts in trying to explain it to me.

[clkorm3]

Question: A turbojet aeroplane has a planned take-off mass of 190 000 kg; the cargo load is distributed as follows: cargo 1: 3 000 kg; (3.50 m from reference point) cargo 4: 7 000 kg. (20.39 m from reference point) Distance from reference point to leading edge: 14m Length of MAC = 4.6m. Once the cargo loading is completed, the crew is informed that the centre of gravity at take-off is located at 38 % MAC (Mean Aerodynamic Cord) which is beyond the limits. The captain decides then to redistribute part of the cargo load between cargo 1 and cargo 4 in order to obtain a new centre of gravity location at 31 % MAC. Following the transfer operation, the new load distribution is:
cargo 1: 6 000 kg; cargo 4: 4 000 kg

Can someone please tell me how this is worked out.

[keith williams]

It is not clear what your question is asking.

Is it asking how much cargo must be moved to move the C of G from 38% MAC to 31% MAC?

Or is it asking where the C of G will be after when there is 6000 kg in Hold 1 and 4000 kg in Hold 4?

Initial condition.
MAC length = 4.6 meters, so 38% MAC is 4.6 x 0.38 = 1.748 meters aft of the MAC leading edge.

The MAC leading edge is 14 meters aft of the datum, so this is 14 + 1.748 = 15.748 meters aft of datum.

Total moment = total mass (190000 kg) x CofG position (15.748 m) = 2992120 kn m.

Final condition.
CofG is at 31% MAC which is 4.6 m x 0.31 = 1.426 meters aft of the MAC leading edge.

The MAC leading edge is 14 meters aft of the datum, so this is 14 + 1.426 = 15.426 meters aft of datum.

Total moment = total mass (190000 kg) x CofG position (15.426 m) = 2930940 kg m.


Calculation of cargo to be moved
Required moment change = new moment (2930940) – Initial moment (2992120) = -61180 kg m.

Moment change = cargo mass moved x distance moved.

Distance moved = new position – initial position

Distance moved = Hold 1 at (3.5 m) – (hold 4 at(20.39 m) = -16.89 meters.

Required moment change = -61180 kg m

Dividing required moment change (-61180 kg m) by distance cargo is moved (-16.89 m) = 3622 kg.

This means that 3622 kg of cargo must be moved from hold 4 to hold 1 to move the C of G from 38% MAC to 31% MAC.


But the final line of your statement of the questions states that “ Following the transfer operation, the new load distribution is: cargo 1: 6 000 kg; cargo 4: 4 000 kg”. If this is correct then only 3000 kg has been moved, so the new C of G will not be at 31% MAC.

Moving 3000 kg a distance of (-16.89 m) give a moment change of 3000 kg x (-16.89 m ) = -50670 kg m.

Adding this to the initial moment gives 2992120 kg m. – 50670 kg m = 2941450 kg m.

Dividing this by the total mass gives a new C of G position of 2941450 kg m / 190000 kg = 15.48 meters.

Subtracting the position of the MC leading edge gives 15.48 – 14 = 1.48 meters.

Dividing this by the MAC length then multiplying by 100% gives 1.48 m / 4.6 m = 0.322, which is 32.2% MAC.

So the new C of G position after moving 3000 kg of cargo is 32.2% MAC.

[clkorm3]

Thank you so much. For breaking it down for me. It's one of the questions on ATPL and one that doesn't also give any explanation too. But thank you.

[taxistaxing]

Does anyone have a useful nemonic/memory technique for remembering the answers to questions such as:

"Turbine blade stages may be cassified as either impulse or reaction. In an impulse turbine stage:

A. The pressure rises across the stator bades and remains constant across the rotor blades;

B. The pressure remains constant across the stator blades and drops across the rotor blades;

C. The pressure drops across the stator blades and remains constant across the rotor blades;

D. The pressure remains constant across the stator blades and rises across the rotor baldes."

(answer in this example is C).

They are a real bugbear of mine...

Tks.

[clkorm3]

This question is very confusing and one that i can not work out.
Although it doesn't state it but I checked cap mrjt1 but even then I don't get the answer it is saying. Can anyone please help

The MRJT 1 CAP STATES:
MS Taxi Mass: 63060
MSTOM 62800
MSLM 54900
MZFM 51300


Question: Prior to departure the medium range twin jet aeroplane is loaded with maximum fuel of 20100 litres at a fuel density (specific gravity) of 0.78. Using the following data - Performance limited take-off mass 67200 kg Performance limited landing mass 54200 kg. Dry Operating Mass 34930 kg. Taxi fuel 250 kg. Trip fuel 9250 kg, Contingency and holding fuel 850 kg, Alternate fuel 700 kg. The maximum permissible traffic load is
Answer: 13090 kg.

[hvogt]

I get 12442 kg max traffic load.

The proposed traffic load of 13090 kg would exceed the max structural take-off mass, so there seems to be an error in the question.

[RichardH]

IF you didn't take into account the MRJT from CAP696/7 and just worked with the given figures then 13092 is the maximum traffic load, but not the clearest question ever produced. Would be better saying "an aircraft" with no reference to MRJT.

20100 * .78 = 15678 kg total fuel - 250 taxi = 15428 kg T/O - 9250 kg trip = 6178 kg on landing.

67200 - 34930 - 15428 = 16842 maximum T/L at take off.
54200 - 34930 - 6178 = 13092 maximum T/L for landing.

Unless the question says "refer to CAP696/7" then I suggest you go with the raw data provided however poor it might be.

[clkorm3]

Thank you for your assistance. Highly appreciated.

[clkorm3]

can someone please tell me how this is worked out.

Question: If an aeroplane performs a steady co-ordinated horizontal turn at a TAS of 200 kt and a turn radius of 2000 m, the load factor (n) will be approximately: 1.1.

Many thanks

[hvogt]

You can use the following two formulae to calculate the bank angle, and then the load factor:

tan phi = v² / (g * R)

phi: bank angle [°]
v: speed [m/s]
g: gravitational acceleration [m/s²] (g = 9.80665 m/s²)
R: radius of turn [m]

n = 1 / cos phi

n: load fator

[Lightning Mate]

I set this question and submitted it to the Dutch SET a few years ago.

The fairly accurate answer is 1.13, which was rounded down to 1.1. http://images.ibsrv.net/ibsrv/res/src:www.pprune.org/get/images/smilies/smile.gif

Why not ask your tutor how to calculate the answer??

[Macarto]

Can anyone explain to me how does being tail heavy, bringing CG to aft of aircraft increase its range?

[Lightning Mate]

It reduces the tailplane downforce so the effective weight of the aeroplane is reduced.

This means that the wing may operate at a slightly reduced angle of attack, thus decreasing induced drag.

[clkorm3]

Question: At reference or see Loading Manual MRJT 1 Figure 4.14. The medium range twin jet transport is scheduled to operate from a departure airfield where conditions limit the take-off mass to 65050 kg. The destination airfield has a performance limited landing mass of 54500 kg. The Dry Operating Mass is 34900 kg. Loading data is as follows - Taxi fuel 350 kg Trip fuel 9250 kg Contingency and final reserve fuel 1100 kg Alternate fuel 1000 kg Traffic load 18600 kg Check the load and ensure that the flight may be operated without exceeding any of the aeroplane limits. Correct answer is:
The flight is 'landing mass' limited and the traffic load must be reduced to 17500 kg.

How is this worked out can some please help and secondly does block fuel include startup and taxi fuel.

[Meikleour]

clkorm3:you come across as a student who has not got a real grasp of your subject. I suggest you get a) a better text book b) or a new tutor rather than trying to get others to do the maths for you!!

None of these questions is difficult and is the sort of calculation that is regularly done on the line.

[keith williams]

The author of your question has specified the MRJT1 aircraft and the loading manual, he/she has then failed to apply some of the limits that are specified in the manual for this aircraft. This error has caused him/her to reach the wrong solution.

The full process is detailed below.

First use the info in the question plus the limits in the CAP696 to determine which limits are the most restrictive to each stage of the flight.

PLTOM = 65050 From question
MSTOM = 62800 From CAP 696

Neither of the above limits may be exceeded, so the Regulated Take-Off Mass RLTOM = 62800

PLLM = 54500 From question
MSLM = 54900 From CAP696

Neither of the above limits may be exceeded, so the Regulated Landing Mass RLLM = 54500

Max Structural Taxi Mass MSTM = 63060 taken from CAP696

Maximum Structural Zero Mass MZFM = 51300 taken from CAP 696.

Then use the most restrictive limits and the load values provided in the question to calculate the mass at each stage of the flight. Compare these masses with the limits to determine whether or not the limits have been exceeded.

Start/Taxi Mass
DOM 34900
Plus Taxi fuel 350
Plus Trip fuel 9250
Plus Reserve fuel 1100
Plus Alternate fuel 1000
Plus Traffic Load 18600
Total = 65200 This figure is 2140 greater than the Maximum Structural Taxi mass of 63060 so the traffic load must be reduced by 2140 to 16460 to remain within the MMSTM.

Take-off Mass
DOM 34900
Plus Trip fuel 9250
Plus Reserve fuel 1100
Plus Alternate fuel 1000
Plus Traffic Load 18600
Total = 64850 This figure is 2050 greater than the RLTOM of 62800 so the traffic load must be reduced by 2050 to 16550 to remain within the RLTOM

Landing Mass
DOM 34900
Plus Reserve fuel 1100
Plus Alternate fuel 1000
Plus Traffic Load 18600
Total = 55600 This figure is 1100 greater than the RLLM of 54500, so the traffic load must be reduced by 1100 to 17500 to remain within the RLLM


Zero Fuel Mass
DOM 34900
Plus Traffic Load 18600
Total = 53500 This figure is 2200 greater than the Maximum Zero Fuel Mass of 51300, so the traffic load must be reduced by 2200 to 14600 to remain within the MZFM limit.

None of the above limits may be exceeded, so the most restrictive condition must be applied. This is the removal of 2200 kg of traffic load to remain within the MZFM.

[Lightning Mate]

Have a great Christmas Keith.:)

[keith williams]

:DAnd you LM.

I see that you have your own thread in the Military forum. Now if that isn't fame I don't know what is! :D :D

[flyer696]

Hi!

First of all I want you to refer to jeppesen approach plates for LKPR (pages 10-9) or UUEE (pages 20-9). There is some interesting change in cycle 1225 and 1226. In old cycle when you will refer to 10-9 table with visibility minumums for NDB approach you will find visibility given in meters. If in front of the value is letter R of course that means that it's RVR. In the cycle 1226 we can find letter C in front of the value. And the letter C is introduced only in the latest database. Can somebody explain me what letter C actually means? Why there is change like that? Any JAR paragraphs?

Looking forward for fast response,

Konrad

[ford cortina]

Flyer696, I am a little confused here.
Both chart numbers given, for Prague and Sheremetyeyo are Taxi Charts.

LKPR try charts 16-1 and 16-2
UUEE try charts 26-1, 26-2, 26-3 and 26-4.

I think you will find what your after there.

[500 above]

Cortina

I'm also confused, guessing he's assuming the 16-1/2's for Prague. I'm wondering whether he's got some kind of a training manual for the ATPL's as I can only see mention of RVR - as expected.

[ford cortina]

Thanks 500, I thought my iPad had gone loopy...:ok:

[flyer696]

My mistake with LKPR. I ment only UUEE chart 20-9S1 where you can find letter C.

Thanks to "500 above" reply I'm now studying difference between CMV, VIS and RVR so thanks a lot for link.

[500 above]

Take a look at EGHG for example, 16-1. You will see both RVR and CMV posted.:ok:

http://www.pprune.org/tech-log/449474-cmv-approach-plates.html

AOM for take-off and landing are either shown on Jeppesen instrument approach or aerodrome charts or on a separate minimums listing. Landing minimums will be shown as RVR, but values above 2000m will be designated as Converted Meteorological Visibility, prefixed “CMV”. Take-off minimums are shown without prefix because they are either RVR or VIS. Circling minimums are always visibilities which is indicated in the circling minimums box. For the separate minimums list- ings RVR, CMV and VIS are abbreviated as R, C and V. The following table is used to convert a reported VIS into RVR/CMV.


CONVERSION OF REPORTED MET VIS TO RVR/CMV

Lighting elements in operation
RVR/CMV = Reported MET VIS x
Day
Night
HIALS and HIRL
1.5
2.0
Any type of lighting installation other than above
1.0
1.5
No lighting
1.0
Not Applicable

From the Jepp EAWM General page 197

[flyer696]

I see ;) So why sometimes they publish CMV instead of VIS? So far I assume that RVR is reported only up to RVR1500 and lets say 1600 will be reported as CMV1600.

So why in EGHG 16-1 in straight-in landing rwy 09 W/o DME we can see CMV 3000m but in circle to land there is VIS 1500 instead of CMV?

And another question according EGHG. What about minimums for airplanes not capable of CDFA (for example C172)? Is it legal to proceed through stepdown DME fixes or do they need to apply minumums only for W/o DME?

[500 above]

Pretty much, if above 2000m it's now a CMV.

Landing minimums will be shown as RVR, but values above 2000m will be designated as Converted Meteorological Visibility, prefixed “CMV”

I think (a while back...) during the ATPL met ground school anything below 1500m was reported as RVR if measured by transmisometers and anything above was a visibility.

I'd just use the non CDFA minimums for a steam gauge spamcan at Yeovil if I was to operate in there, but I just used it as an example.

[flyer696]

Ok so below 2000 we have RVR and above there will be CMV. What about VIS? Only in circle to land and airfields with no RVR equipment and values below 2000?

[500 above]

This also has to be in mind:

DEPICTION OF EU-OPS AOM IN CASE OF EXISTING STATE MINIMUMS

If State minimums are officially published, the depiction of AOM may differ from the standard depiction where all values are expressed as RVR or CMV.
a. If RVR/CMV and VIS are charted together, the RVR value is compulsory. If no RVR is repor- ted, the VIS has to be used without conversion.
b. No prefix is charted if RVR/CMV and VIS is identical. The reported RVR is compulsory. If no RVR is reported, the VIS has to be used without conversion.
c. If only VIS is charted, the VIS has to be used without conversion.

Jepp page 206 EAWM, General.

I see So why sometimes they publish CMV instead of VIS? So far I assume that RVR is reported only up to RVR1500 and lets say 1600 will be reported as CMV1600.

Just to clarify, you will only be passed an RVR or Vis. You, the crew, calculate a CMV using the tables.

[flyer696]

Thanks a lot for making me all that stuff clear ;)

[Macarto]

Could someone explain to me why do we use a Voltmeter to measure output for a generator but a ammeter / loadmeter to measure output for an Alternator?

[mad_jock]

Its all to do with the output.

I suspect you haven't got the basic concepts abouts whats going on.

All About Circuits : Free Electric Circuits Textbooks (http://www.allaboutcircuits.com/)

Try reading through that a bit and see if you can work it out yourself.

If you really get stuck just ask.

Power factor is the interesting bit :ok:

[Skymaster15L]

Happy New Year to Everyone.

Have been studying the limitations chapter in POF and in particular, the V-n diagram that deals with the load factor limits of the flight envelope, and the various important speeds that you find on that envelope.

Comparing the definitions of both Design maneuvering speed (Va) and Max Structural cruising speed (Vno), I am a bit confused. I would like to know what exactly is Vno useful for? I understand it is the limit above which flight should be attempted in smooth air only (no turbulence?) and then only with caution. Why is it then that everywhere else I read, it says that when encountering turbulence you should slow down below maneuvering speed, in addition to when you are engaging in maneuvers, etc.

I am aware that the definition of Va is often a varied one and commonly misunderstood by many pilots, usually along the lines that you think you are guaranteed in throwing unlimited abuse at an airplane thru abrupt control deflections, etc without having it fail as long as you are below Va, so I am not inviting to dissect this subject unless necessary in answering this post.

So based on the definitions from aviation textbooks of what each speed is, I'd simply like to know which one should be used for what then and why? I too would think Va is the more important one, since it deals with protecting the airframes structural integrity, but then I also been in many an airliner that when it has experienced turbulence, I've never perceived it to slow down a substantial amount as if it was aiming for a speed below Va. So then, in which circumstances does Vno has a practical application?

Thanks

[Haroon]

hi need some help in the following two questions:

Q.1. If you correctly tuned in a VOR situated to your east, your RMI should read ___ and your OBS would read ___

a) 000; 000 with needle central and TO indicated
b) 090; 090 with needle central and FROM indicated
c) 000; 000 with needle central and FROM indicated
d) 090; 090 with needle central and TO indicated

Just cant visualize this one.

Q.2. The accuracy of a DME:

a) is approximately ±0.5nm
b) decreases with increase of range
c) increases with increase of altitude
d) is approximately ±2 nm

Which accuracy is this? Slant range accuracy increases with increase in range.

Thanks

[RTN11]

a) 000; 000 with needle central and TO indicated
b) 090; 090 with needle central and FROM indicated
c) 000; 000 with needle central and FROM indicated
d) 090; 090 with needle central and TO indicated

Just cant visualize this one.

Q.2. The accuracy of a DME:

a) is approximately ±0.5nm
b) decreases with increase of range
c) increases with increase of altitude
d) is approximately ±2 nm

1) you are tracking away from the VOR, therefore have 000 at the top, with the needle centered and the From flag. It would be incorrect if it were tracking 000, with 180 on the VOR and a FROM flag, as you would be going to the VOR and indications would be in reverse sense.

2) the error decreases with range. If you are over the station at 6000' (1nm) the DME will read 1nm, even though you are at the station. At 50 miles, at 6000', the error will be far less.

[Haroon]

1) you are tracking away from the VOR, therefore have 000 at the top, with the needle centered and the From flag. It would be incorrect if it were tracking 000, with 180 on the VOR and a FROM flag, as you would be going to the VOR and indications would be in reverse sense.

but in this case the VOR will be south of you, whereas the question says its situated to your east.

2) the error decreases with range. If you are over the station at 6000' (1nm) the DME will read 1nm, even though you are at the station. At 50 miles, at 6000', the error will be far less.

Yes thats why posted the question. It says accuracy decreases with increase in range.

[BlueSkyLife]

For Question 1:

Imagine you are VFR for a moment. If you were to see out your window that the VOR is East of your position, then that simply means you are west of it and to (TO) get there you should fly east. Therefore, your RMI will point east as the head of the needle points to the station while the tail gives your position.

A FROM on your OBS, always think of it as your position in relation to the beacon/reference. A TO indication means you should fly East to get to the VOR, FROM means you are LOCATION about of the VOR. If your were to twist the OBS until it shows FROM and then you fly to the VOR your old conventional VOR receiver would be Reverse sensed. HSIs don't sensed revered.

Q2

DME (Distant measuring equipement) not a GPS.... is measured in slant range. The closer you get the larger the error as your aircraft height is more of an issue. So RNT11 dude hit the nail on the head for you answer.

[taxistaxing]

a) 000; 000 with needle central and TO indicated
b) 090; 090 with needle central and FROM indicated
c) 000; 000 with needle central and FROM indicated
d) 090; 090 with needle central and TO indicated

Just cant visualize this one.

Q.2. The accuracy of a DME:

a) is approximately ±0.5nm
b) decreases with increase of range
c) increases with increase of altitude
d) is approximately ±2 nm

Also interested in these questions.

I'm struggling to understand the first I'm afraid. Surely if the VOR is East of your position you will be on the 270 radial and would need a 090/270 OBS selection to get "to" or "from", respectively. Why would you ever get a reading of 000? Is this talking about a relative bearing?

What does 'correctly tuned' mean. Are you tracking inbound or outbound to the beacon?

Second question, if we know the required accuracy of a DME is 1/4nm + 1.25%, we know A and D are wrong.

C is wrong (accuracy decreases with increase in altitude as the vertical 'slant' component increases), so we are left with B as the least worst option. It's true to say slant accuracy increases with range but then DME is line of sight so eventually accuracy will decrease with an increase in range, as you pass beyond line of sight of the transmitter.

[Haroon]

Surely if the VOR is East of your position you will be on the 270 radial and would need a 090/270 OBS selection to get "to" or "from", respectively. Why would you ever get a reading of 000?

thats precisely why i posted the question

It's true to say slant accuracy increases with range but then DME is line of sight so eventually accuracy will decrease with an increase in range, as you pass beyond line of sight of the transmitter.

i was also thinking on these terms but not quite sure if this is really the reason.

[bayblade]

Is it possible that the question Q1. is incorrect?

[Haroon]

i think it is!

how are your ground classes going :cool:

[hvogt]

In my opinion the correct answer to the first question would be d). An RMI has no TO/FROM indication, so if the station is to your east, the needle will indicate east. As for the TO/FROM indication, it should indicate TO with 090 selected on the OBS.

[Haroon]

thats right, i think the question is wrongly marked

thanks everyone for your feedback :ok:

[bayblade]

going good by God's grace:)

[Haroon]

Any idea about this one?

What is the PRF given 50 micro second pulse width and a range of 30 nm:

1620 pps
810 pps
3240 pps
3086 pps

What I know is:

Max Theoretical Range = c / 2 x PRF

30 = c/2prf

prf = 2700

Am i missing something, like something to do with 50 micro second pulse width or there is something else to it?

thanks

[RichardH]

Long time since I did this but I THINK its 81000/50 = 1620 but can't remember why!

[taxistaxing]

That's a strange one.

Using 162000 nm/sec as SOL I make D the closest. A seems to equal 50 nm not 30.

[Haroon]

i think there's a typo in the question. It must be 50 nm

[NoJoy]

This was great help thanks a lot guys!

[paco]

There are 96 articles and 18 Annexes, of which only about 6 are questioned on. Facilitation?

You are not expected to know the fine details of contents of the articles, but definitions (sovereignty, etc) would be worth swotting up on.

Also customs requirements and the 5 (out of the 9) freeedoms.

Phil

[mad_jock]

In real life its Facilitation and Dangerous goods your most likely to come across.

[Haroon]

Hi

Q.1. When switching on the weather radar, after start-up a single very bright line appears on the screen. This means that the:

scanner is not rotating
scanning of the cathode ray tube is faulty

Q.2. On switching on the AWR a single line appears on the display. This means that:

the CRT is not scanning
the antenna is not scanning

Whats the main difference between the two?

[Lightning Mate]

Long time since I did this but I THINK its 81000/50 = 1620 but can't remember
why!

Sounds like you might be one of my ex-students.

The 81000 is one million divided by 12.36, the latter being the "radar mile" :)

[keith williams]

Q.1. When switching on the weather radar, after start-up a single very bright line appears on the screen. This means that the:

scanner is not rotating
scanning of the cathode ray tube is faulty

Q.2. On switching on the AWR a single line appears on the display. This means that:

the CRT is not scanning
the antenna is not scanning

Whats the main difference between the two?

The main difference appears to be the use of the words "very bright" in question one. If the electron gun in the CRT is repeatedly scanning either vertically or horizontally, instead of both vertically and horizontally, this will concentrate the energy in a single line. This concentration of energy is likely to make the line very bright. In this case the most accurate answer would be "scanning of the cathode ray tube is faulty".

For question 2, if the CRT is working correctly but the antenna is not scanning, then only the radar returns from the area in line with the scanner would be displayed. Whether or not these would produce a very bright line would depend upon what returns were being received from that area.

[Haroon]

Hello

Lightning Mate: Can you pls explain a little bit how to solve that question?

Keith: Thankyou!

[chr15ru55e11]

Hey,

I recently heard a few questions asked at some interviews but cant find the answer anywhere.

The question was, by law how long must a battery last if power is lost. I have a feeling its 30 mins but I cant be sure. I have checked EU ops and cant find the answer anywhere. Anyone know what it is for a fact?

[joker737]

On the Boeing 737-800 it's 60 mins in a good world... And that's using both the main battery and alternate battery!

[chr15ru55e11]

Thanks for the reply. Do you know if that is the legal limit of one hour (on a good day) or is that Boeing's own choice to make it last longer?

[deben_]

Hello Everyone.

I am looking for regular or part time ground school for CPL(H) Exam. Is there any in the UK and how much it cost?

Any suggestion is much appreciated.

Cheers

[Haroon]

How does one do this one:

If a radar pulse contains 300 cycles of RF energy at a frequency of 600 MHz, the physical length of the pulse is:

1550 metres
150 metres
1.5 metres
0.15 metres

thanks

[paco]

"Q.1. When switching on the weather radar, after start-up a single very bright line appears on the screen. This means that the:

scanner is not rotating
scanning of the cathode ray tube is faulty"

Interesting - as one who used to service monitors my first port of call would be the power supply :)

Deben - check out the big ad on the right