@37thousandfeet
Its the DGCA R.K. Puram Address.!

@Aviator09
thanks...

It has been decided that Pilot Licence examination - July 2011 session for ATPL
Category shall be conducted only at Delhi, Mumbai, Kolkata and Chennai.



http://dgca.nic.in/public_notice/PN%...pplication.pdf

Hi parasite tango , if possible can u tell us till wht depth does thy ask Qs abt FMS,INS,IRS,ECAM and EICAS, AUTOLANDING system

Radius of cone of confusion = A/C altitude (in NM) X Tan 50

and then radius X 2 to get the diameter of cone of confusion in NM. :ok:

a question to anybody who is willing.... i have a current CPL/MEIR..... i passed air nav( composite) and regs exam (CPLCG) what papers do i need for my ATPL..... i know radio navigation is one, any other i should know about ?

also can i be issued with ATPL(frozen ) with the pass results of ATPL papers which are valid for 5 years ? ( if i do no flying at all that is )

Mike Sierra - I think you can work it out this way.

Your descent gradient is 12% which is 0.12 times the groundspeed.

0.12 * 540 kts = 64.8 kts (Now, 1 knot = 101.33 feet/min)

So your ROD works out to be = 64.8 * 101.33 = 6566 feet/min

Mike Sierra - The answer to you second question could be worked out using the 1 in 60 rule.

Percentage gradient is the tangent of 5.2 deg * 100 = 2.977 deg

Height of the aircraft = 7/2.977/60 ( 1 in 60 rule) = 0.3473 Nm

Nm to feet conversion = 0.3473 * 6080 = 2111.58 feet

I hope it's correct.

As with all such problems it is best to try to get an understanding of what the information means.

A 12% gradient means that your vertical speed is 12% of your horizontal speed.

12% of 540 knots = 540 x 12 / 100 = 64.8 knots.

1 knot = 101.33 ft/min so 64.8 knots = 64.8 x 101.33 = 6566.184 ft/min.

Smurf84
Your method can be used but you made an error in converting the tangent into an angle.

Percen% gradient = 100 x tangent so for 12% the tangent is 0.12 as you have said.

But for a tangent of 0.12 the angle is 6.843 degrees.

Dividing this by 60 and multiplying by 540 knots gives 61.58 knots.

Multiplying this by 101.33 ft/min/knot gives 6240.40 ft/min.

Thanks for the correction, Keith. I reckon I'll have to go through it again.

Doubts
 

The below are some questions from Oxford instrumentation book - objective revision 4 and 5.

Q1) An a/c is descending at a contant mach number. If the a/c is descending through an inversion layer, the RAS will:

a)remain constant
b)increase
c)decrease
d)decrease then decrease more slowly

Answer is b) increase.
however during descent in inversion, temp decreases at lower altitude. so, LSS decreases. since M.no = (TAS / LSS), therefore TAS should also decrease, in order for M.No to remain constant.
Yes, density increases at lower altitudes.
Now, RAS=(1/2)*density*(TAS squared).
Therefore effect of decreasing TAS is more, then increase of density.
so answer should be c) decrease.
Can anyone explain?

Q2)A perfectly frictionless DI, is corrected to give zero drift on the ground at 30 degree north. The DI is set to read 100 degree in an a/c stationary on the ground in latitude 45 degree north. The reading after 45 mins will be:

a) 92.33 degree
b)102.50 degree
c)97.78 degree
d) 103.75 degree

Q3) with constant drift during flight, the a/c heading will:

a)increase by more than 10 degree
b)decrease by less than 10 degree
c)increase by less than 10 degree
d)remain constant

Answer is b). what;s the logic?

Quote:

---

Q1) An a/c is descending at a contant mach number. If the a/c is descending through an inversion layer, the RAS will:

a)remain constant
b)increase
c)decrease
d)decrease then decrease more slowly

Answer is b) increase.
however during descent in inversion, temp decreases at lower altitude. so, LSS decreases. since M.no = (TAS / LSS), therefore TAS should also decrease, in order for M.No to remain constant.
Yes, density increases at lower altitudes.
Now, RAS=(1/2)*density*(TAS squared).
Therefore effect of decreasing TAS is more, then increase of density.
so answer should be c) decrease.
Can anyone explain?

---

You are making the mistake of assuming that the rate of decrease in TAS squared is greater than the rate of increase in air density (which determines the RAS at any given TAS.

LSS is proportional to the square root of absolute temperature so as we descend at constant Mach in an inversion, the decreasing temperature will cause the LSS and our TAS to decrease.

But as we descend both the increasing static pressure and the decreasing air temperature will cause the air density to increase. This will cause the RAS and EAS at any given TAS to increase.

The effect of the increasing density will be greater than the effect of the decreasing LSS so the overall effect will be that the EAS and RAS both increase while the TAS decreases at constant Mach.

You should be familiar with the simple ERTM graphs, which will enable you to answer this type of question in an exam without taking the time to go through all of the above arguments. If you do not have them please send me a PM and I will send them to you.

Quote:

---

Q2)A perfectly frictionless DI, is corrected to give zero drift on the ground at 30 degree north. The DI is set to read 100 degree in an a/c stationary on the ground in latitude 45 degree north. The reading after 45 mins will be:

a) 92.33 degree
b)102.50 degree
c)97.78 degree
d) 103.75 degree

---

If you cannot answer this question then you need to study gyro errors before taking the exams.

In the northern hemisphere Earth Rate drift rate in degrees per hour is

Earth Rate in degrees per hour = -15 x Sin Latitude

The latitude net is designed to create a drift rate in the opposite direction to compensate for Earth Rate.

In the northern hemisphere Latitude Nut Rate drift rate in degrees per hour is

Lat Nut drift rate in degrees per hour = + 15 x Sin Lat Nut Setting

Adding the two together gives the total drift rate, which is

15 x (Sin lat nut setting – Sin latitude)

Multiplying this by the time gives the total drift.

Inserting the data from the question gives.

The gyro is said to be totally frictionless so there will be no random wander

So Total drift =( (+15 x Sin 30) + (-15 x Sin 45) ) x 0.75 hours

Total drift = -2.32995 degrees

Adding this to the initial heading of 100 degrees gives

100 – 2.32995 = 97.67 degrees.

The closest option to this is option c.


[QUOTE]
Q3) with constant drift during flight, the a/c heading will:

a)increase by more than 10 degree
b)decrease by less than 10 degree
c)increase by less than 10 degree
d)remain constant

Answer is b). what;s

As it stands this question makes no sense. But if we assume that it is the same aircraft as in the previous question, then because it is in the northern hemisphere the earth rate drift will always be a minus value. This will cause the heading to decrease. Only option b includes a decrease, so we do not need to look any further. But the question states that the aircraft is “in flight” so we really need to consider Transport wander. But the question does not give us enough information to do this.

Please help
 

Hey Guys
I am appearing for ATPL for the first time. Could some1 please provide some notes on
1.Air Data Computer
2.Auto pilot
-autoland, sequence of operations
-system concept of autoland, go around, take off, fail passive, fail operational
redundant

Would really appreciate!:)

@ keith williams. Check ur PM
 

Could anyone clarify about Mode 6 and 7 in EGPWS.

Mode 7 is for windshear alert below 1500 feet and gives windshear warning both during approach and takeoff.

similarly, Mode 6 gives bankangle warning. Is the bank angle warning only during approach or also during t/o?

If you do a GOOGLE search for EGPWS you will find a lot of links inlcuding some to the HONEYWELL website.

These links cover most of what you will need to know about the subject.

Earlier question
 

I skipped a vital sentence from one of my earlier questions by mistake. I apologise. Here is the full question:

An INS equipped a/c flies from 56 N 20 W (waypoint 3) to 56 N 30 W (waypoint 4). With constant drift during flight the a/c heading will:

a) increase by more than 10 degree
b) decrease by less than 10 degree
c) increase by less than 10 degree
d) remain constant

Answer: b (How less than 10 degree?)

My explanation: Departure=335.516 NM

ERW comes to -12.435 degrees/ hr

TW = GS /60 * tan lat

How to calculate GS ?

With that extra paragraph it is clear that this is not a gyro question. It is a great circle question (INS systems fly great circles).

It is asking for the amount by which the great circle track changes from A to B.

If you have not done these before I suggest you read up on Great Cricles and Conversion Angles. Then have a go at it.

When you post yourmethod and answer we can comment on it.

This is what I can make sense of. Unable to think beyond it.

Conversion angle=0.5*dlong *sin mean lat
=0.5*10*sin56
=4.145 degree(or 4 degrees)

so, initial GC track at waypoint 3(or 20 W)=(270+4)= 274 degrees

Now drift = angular difference between track and heading

HDG= 270 degrees (therfore drift= 4 degrees SB)

Similarly GC track at waypoint 4 (or 30 W) = (270-4)= 266 degrees

Here drift = 4 degrees port (assuming hdg remains 270)

i think the answer has to be "B"..
the heading is decreasing by approx 8.6 degrees..

cause when u depart from way point 3 your hdg is 274.3..
when you arrive at way point 4 your heading is 265.6 ...
so with constant drift throughout the flight...your hdd is decreasing by less than 10 degrees....i.e. 8.6...

please correct me if i am wrong...

Your method and answer are correct Wings of fire.

But your comment that

Is true when discussing drift in the normal sense, but this question isn't really about drift. I think that the examiner put in the words "constant drift" just to eliminate any unwanted complications. It would probably have been better to have said "in nil wind conditions".

The difference between the Great Circle track and the Rhumb Line track isn't realy an example of drift. At the start of this leg when the heading is 274.145, the track will also be 274.145 (assuming nil wind). And at the end of the leg when the heading was 265.855 the track will also be 265.855. But the average track over the whole leg would have been 270.

Thank you keith. I appreciate your sincere efforts.

ATPL on 19,20 and 21........all the best guys.....

guys! is there gonna be a list of admitted candidates any sooner?
or a VENUE to where to take the exam! Its already the 16th and 19th is the exam and no Venue and no admitted list and no nothing! OmG!

me too, waiting for the list and Venue! whats up with DGCA being so slow in declaring the listed candidates. Hardly any reaction time for those who dont make it in the list.

Well, Thats DGCA, Everything is fair in case of DGCA. The are the Brand-ambassadors of Laziness. :ugh:

@keith williams
 

Hello keith. Can u solve this question. Again from radio aids oxford book- vor chapter.


Q4)what is the theoretical maximum range that an aircraft at FL 420 will obtain from a vor beacon situated at 400 feet above msl?
a) 225 nm
b) 256 nm
c) 281 nm
d) 257 nm

answer is a) 225 NM

My explanation:

Range = 1.25*(Root of ht + root of hr)

=1.25*(root of 400 + root of 42000)

= 281 NM

Your soluton is correct.

The author appears to have forgotten to multiply by 1.25.

This kind of problem would be avoided if the book included worked solutions for each question, but this obviously involves more work for the author.

do all of you who applied,have your name in the admitted candidates list?

i dont understand, my form , draft were as per guidelines but still i didnt get my roll no.:ugh::ugh::ugh::ugh:

and they are saying

Application of those candidates whose name does not appear in the
above list has been rejected either due to incomplete application/not enclosing
the required documents/wrong demand draft. CEO shall entertain no
communication from the candidates, whose applications have been rejected.

What to do now????:confused::confused:

So the three venues of Delhi are!

JASOLA, DWARKA & ROHINI! Looks like DGCA is on a Pilots DILLI DARSHAN Mission!

No idea why such absurd locations were chosen! All three are in Different corners of the City!

One Centralized Location with different Timings for Exams over 3-4 days could have been done! Would have been easier for people not familiar with the City coming from outside!

On top of that they post the list of names just 2 days before the exam!
Don't want to entertain any issues of names they haven't put in the list!

Dictatorship General of Civil Aviation, India! :ugh: :ouch:

Next punch would be watching the systems crash!

same here buddy...the clowns havent even posted the rejection list so one doesnt even know what went wrong :mad:

@ Roll No.
 

Does anybody have a idea that do we have to go for stamping from DGCA official as in CPL exam or just a print out of roll no. will work..

Please Revert

The great circle track X - Y measured at x is 319°, and Y 325° Consider the following statements:
a Southern hemisphere, Rhumb line track is 322°
b Northern hemisphere, Rhumb line track is 313°
c Southern hemisphere, Rhumb line track is 331°
d Northern hemisphere, Rhumb line track is 322°


A great circle track joins position A (59°S 141°W) and B (61°S 148°W). What is the difference between the great circle track at A and B?
a It increases by 6°
b It decreases by 6°
c It increases by 3°
d It decreases by 3°


Thanks

a Southern hemisphere, Rhumb line track is 322°

b It decreases by 6°

Damn i still remember :D:D

All the best guys for your ATPL:ok::ok:

The first ones right.

But the second ans. is given as Increase by 6

When taking the exam the most important thing is to select the correct answer.

When preparing to take the exam the most important thing is to learn the correct method.


The important thing to remember here is because both the rhumb line and the great circle join the two points, the rhumb line track is the average of the great circle track at the two points. So in this question we just need to add the two together and divide by two to get (319 + 325) / 2 = 322.

Great circles are always convex to the nearest pole.

We have a great circle that is changing from 319 to 325, so we must be in the southern hemisphere. Sketch the situation if have difficulty visualising it.


For this one we need to use the conversion angle formula

The conversion angle is the difference between the great circle and the rhumb line.

Conversion angle = 1/2 x Sin mean latitude x difference in longitude.

In this case we have

Mean latitude = (59S + 61S) / 2 = 60S.

So Conversion angle = 1/2 x Sin60S x ( 148W - 141W)

If we use the conventions that

Latitudes North are positive
Latitudes South are negative
Differences in longitude to the West are positive
Differences in longitude East are negative

We get

Conversion angle = 1/2 x Sin -60 x 7W = -3.03 degrees.

The conversion angle is the difference between the great circle and the rhumb line so at position A the great heading is 3.03 degrees less than the rhumb line.

But the great circle and the rhumb line meet at positioon B so the great circle heading must gradually increase until it becomes 3.03 degrees greater than the rhumb line.

Starting 3.03 less and ending up 3.03 more, the great circle must have increased by 6.06 degrees.

If the use of plus and minus values is too confusing, then simply sketch the situation.

Great circles are always convex to the nearest pole.

We are in the southern hemisphere so the great circle initialy heads south of the starting point then gradually curves to the north until it reaches the second position. We are going west so this curving must represent a gradual increase in heading.

So the answer to this question is increasing by 6.06 degrees.

Thank you sir.

Truly genius. :D:D:D

Please also could you clarify.

On a Direct Mercator chart at latitude of 45°N, a certain length represents a distance of 90 NM on the earth. The same length on the chart will represent on the earth, at latitude 30°N, a distance of :
a 45 NM
b 73.5 NM
c 78 NM
d 110 NM

&


The standard parallels of a Lambert's conical orthomorphic projection are 07°40'N and 38°20' N. The constant of the cone for this chart is:
a 0.60
b 0.39
c 0.92
d 0.42

@azax
Dude seriously .. I would suggest put some effort..and you will get it...

Good luck to all appearing for ATPL

@ i did get the 1st one later..
cos30/cos45 * 90 nm = 110.2 nm

but i have no clue of 'constant of cone'
if you do know..you might as well post the solution