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-   -   Indigo Call letters for Freshers (https://www.pprune.org/south-asia-far-east-wannabes/434171-indigo-call-letters-freshers.html)

yash 3rd Aug 2011 05:59

@flyhigh27

After the CASS assessement you have your GD and Final Interview all in one day .

LieDetector 3rd Aug 2011 06:10

all the best guys!

This was the last written for this year, next will be conducted by CAE early next year.

lesternoronha 3rd Aug 2011 06:21

@LieDetector
 
are u sure about this exam being the last this yr......that will really suck...all the best to all the guys who cleared it....make the best of it!!!:ok:

Uniform. 3rd Aug 2011 06:24

@lesternoronha

WHT??? :ooh:

may i knw from where did u get this info.?

LieDetector 3rd Aug 2011 06:32

yes from now on the exam will be every 3 months starting end of this year or early next.

lesternoronha 3rd Aug 2011 06:39

well they have that 3 month cool off period anyways....waited so many yrs whats a few more months??

9FLYJET9 3rd Aug 2011 07:27

Dude who told you that there will not be an an exam in August.
Are you sure about it.
Hope that is not true.
Please.

sukhoi30mki 3rd Aug 2011 07:36

Did anyone get confirmation from CAE that there will be no more exams conducted this year, guys going for CASS please confirm

9FLYJET9 3rd Aug 2011 07:42

@ lie detector: who told you that this was the last for this year ? please post the reply asap.would appreciate the help.

wings of fire 3rd Aug 2011 10:20

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?

HuNg_sTaRt 3rd Aug 2011 11:15

TAS = CAS + DE (density error) and Mach no. = TAS/LSS

Mach no. is constant (given). So TAS is constant.

Now, since the a/c is descending through inversion layer the temp will decrease an so will density error and to keep TAS constant(by d fromula)....CAS will increase!!!:ok::ok::ok:

747dash8 3rd Aug 2011 15:25

I had appeared for the june assessment.....so according to the 3 months cool off period.....when am i applicable again ?
September or October ?

And another question.......Is it mandatory to have ME IR current for appearing in the exam ?

aviatepilot 3rd Aug 2011 15:32

dude find out first if the exams are gonna be conducted or not.....As quoted by someone the exams r gonna be now next yr only....

wings of fire 3rd Aug 2011 16:05

@hung_start
 
Quote: Mach no. is constant (given). So TAS is constant.

LSS=38.95*root of temp(in kelvin)

since M.no = TAS / LSS , Therefore LSS changes with decreasing temp. Hence TAS also changes . It can't be constant.

Any one else can explain.

747dash8 3rd Aug 2011 16:24

I read that post by LieDetector.......but since there is no such announcement on either Indigo or CAE's website.......m trying to think positive....
Kindly clear my doubts in this matter if you can.....

Aviator09 3rd Aug 2011 17:44

wings of fire

Q2. Find the real drift due to mechanical imperfections @ 30 N which comes to be 7.5 degree/hour increasing.
total drift=Real Drift + Apparent Drift i.e. 0= RD + 15 sin 30
0= 7.5/hr Increasing + 7.5/hr dec

Now, this gyro is taken to 45N, RD will remain same, no matter what,
Hence, Total Drift= 7.5/hr inc + 15 sin45
= 3.10 degree/hr decreasing

gyro stays here for 45 minutes, so 3.10/hr when computed for 45 minutes comes to be,2.32 Degree/ hour decreasing.

Alteration to the hdg. will be 100-2.32=97.67 degree. (c)

Is the answer correct??

AAC2011 3rd Aug 2011 18:56

as the temp decrease the local speen of sound reduce
you see the inversion the lss decrease as the a/c descend
Now comes the basic maths , as the denominater decrease to keep the full factor constant (mac no.) we have to increase the TAS and in attampt to increase tas ras increase




ahy update for the indigo coming exam , please inform


Q2, 3 do you have the answer and its justification

fly.low 3rd Aug 2011 19:07

has anybody been called for cass yet, who passed on 29th july,2011?

DJ Flyboy 3rd Aug 2011 19:36


Now comes the basic maths , as the denominater decrease to keep the full factor constant (mac no.) we have to increase the TAS and in attampt to increase tas ras increase
You may want to reconsider that..if the denominator decreases..the numerator also has to decrease to keep the fraction constant..here let me make it simple (not that its complicated!).. consider..20/10=2
now if you reduce 10 to 5 then you also have to reduce 20 to 10, 10/5=2

:8:p:ok:

Now for the answer..cas does increase according to an online flt computer..but dont ask me how..stumped on this one..can anybody explain why cas increases while descending through an inversion layer with constant mach?

And the most basic question..is there going to be an August exam???

johney_flying_high 3rd Aug 2011 21:43

A warm hello to every1 reading this...
Could someone please provide me the solution of the following question:

What is the distance in kilometres from 49° S 180° E/W to 58° S 180º E/W?:confused:


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