@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.