@ecks
 

This could be possible but there is one guy named "FlYINGW" on page 58, who did mention that he is from Aug2010 batch and has received the date for GD...
Lets see....Quite possible that we receive the dates today...:ok:

some of my friends from aug 2010 batch got the dates for G.D. but some of them are yet to get it. I hope everyone will get the dates by next 2 days. :ok:

Any one got 25th?

hey guys i am also from last year aug 2010 and still awaiting call from Jet.
will keep all posted. u guys can call me on 9869315394 amar

i also cleared written in august and hvnt got the mail yet :ugh:

heyy all aug 2010 batch guys whom have got dates please tell if you are from MUMBAI

I am from last august and my gd date is 25. I am from mumbai but i know 2 -3 people who are all from mumbai but haven't got the call yet.

But i guess all the people from Feb2011 have received there respective dates...correct me if i am wrong!!!

M frm Aug2010 batch, havent received any e-mail yet

Does any one know when is jet posting for new vacancies???

Doubts
 

Hey guys .could u please help me in solving the below questions. These r from oxford radio aids- vor chapter


Q1) Assuming the maximum likely error in vor to be 5.5 degree, what is the maximum distance apart that beacons can be situated on the centreline of the u.k airwayin order that an aircraft can guarantee remaining within the airway boundary?
a) 54.5 Nm
b) 109 NM
c) 66 Nm
d) 132 Nm

Q2) An aircraft is homing towards VOR which marks the centreline of an airway. The beacon is 100 Nm distant. If the pilot had the airway QDM set on the OBS what deflection of the deviation indicator would be given if the aircraft was on the boundary of the airway? Assume that 1 dot equals 2 degrees.
a) 3 dots
b) 2 dots
c) 2.5 dots
d) 1.5 dots

Q3) Assuming an accuracy of + / - 5.5 degree, the maximum spacing of 2 vor’s defining the centreline of an airway, that would ensure no aircraft could be outside the airway is;
a) 54 Nm
b) 51 nm
c) 109 nm
d) 132 nm

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

So no one from Aug 2010 GD rejected lot has been intimated as of now.....do any one knows about the fate of those who got kicked out in Interview of Aug 2010?
Looks like some got a raw deal while others got a sweet one.:(

some(Aug 2010) did get a call and a few of them are from delhi and chennai as well.

Hey guys, I just read the last few pages and just to help you all clear the doubts ...I am from Mumbai from the august 2010 batch and got my call for the 25th on Friday the 29th itself .. Dont worry everyone will get it soon ..
best of luck

any more call letters on 1st or 2nd???

nothing till now!

Doubts of wings of fire
 

Q1) Apply 1 in 60 rule. 5.5=[(5x60)/d]
where 5.5 = error in vor
5=airway width in uk
d=half the max. distance of the 2 beacons.
Solving which we get d=54.5nm.
Therefore max. spacing=2d=109nm.

Q2) Apply 1 in 60 rule again.
Error=[(5x60)/100]=3 degrees.
Where error is the angle deviated from centreline, the aircraft being on the air way boundary which is 5nm i.e. airway width in uk.
100 is the distance of the beacon from the aircraft.
Given or even generally known that in OBS 1 dot is 2 degrees.
Therefore 3 degrees equals 1.5 dots.

Q3) Same as Q1)

Q4) Range=1.25(Square root of h1 + Square root of h2)
h1 is height of transmitter above msl in feet.
h2 is height of the aircraft above msl in feet.
Hence Range=1.25(Square root of 400 + Square root of 42000)=281nm.

Call letters???
 

If anybody has tried calling the jet airways office,please inform !!!
I tried calling but no response...

@ btommy
 

Thanks.

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?

The CAS/TAS variation due to density error is much greater than the LSS change due to temp. variations.

So when descending in an Inversion layer the LSS in decreasing due to reduced temp and the TAS is also decreasing with CAS const. but the TAS reduction is at a much greater rate than the LSS because of the additional increase in density due to reducing temp. So if you kept CAS const. the TAS will reduce at a much greater rate than LSS and the Mno. will reduce. So to keep Mno. cont. the rate of reduction of TAS has to be reduced by increasing CAS.