Any1 from Aug 2010 batch gt any mails 2day?? Hve u tried calling Jet Airways HR depart???

@flyjet787
 

Quote: So if you kept CAS const. the TAS will reduce at a much greater rate than LSS and the Mno. will reduce.

Buddy , u r assuming , that if we keep cas constant. however it is given in th question that we r descending at a constant Mach number.( Hence TAS will be taken care of due to additional increase of density due to decreasing temp.)

any other opinions to my question?

If you want to keep Mno.const. you will have to control the equation TAS/LSS. How do u control TAS? With CAS.

In normal conditions when you descend keeping Mno. constant you will increase the CAS to increase the TAS so that it matches with the increase in LSS to keep Mno. cosnt.

But when you descend in an Inversion layer keeping Mno. const.with the LSS decreasing the TAS needs to decrease at the same rate as LSS to keep Mno. const. But since the reduction in TAS is at a higher rate (due to the additional density increase due to reduced temp.) compared to the reduction in LSS the Mno. will decrease. So to keep Mno. const. TAS needs to reduce at a slower rate (equal to LSS) for which the CAS has to increase. If you reduced CAS the TAS would reduce at a much greater rate than LSS and the Mno.would reduce.

The additional increase in density will further reduce the TAS and therefore the TAS will actually start reducing at a much faster rate than the LSS (The CAS/TAS variation due to density error is much greater than the LSS change due to temp. variations.)

@flyjet787
 

allright I will take your explanation. Thanks.

Any explanations for remaining 2 questions.

Oxford ATPL
 

Hi All,

Does anybody have the answer key to the Instrumentation book from Oxford aviation services ATPL?
kindly pass on if you do.

thankyou,
P

@wings of fire

@ 30 N -> No Drift
i.e. ER Drift + Real Drift = 0
15Sin30 (Dec(Northern Hemi)) + RD =0
7.5(Dec) + RD = 0
RD = 7.5 Deg/Hr (Inc)

@ 45 N
Total Drift = 15Sin45(Dec) + 7.5 (Inc)
= 10.606(Dec) + 7.5 (Inc)
= 3.1066 Deg/Hr (Dec)

Reading after 45 mins
3.1066 * (45/60) = 2.32995 Deg/45 Mins (Dec)

At 100 degs = 100 - 2.32995 = 97.67005 Degs

Answer (c)

I hope all these oxford questions have nothing to do with the ongoing Jet airways recruitment process?!
And does anyone know how many vacancies Jet is looking to fill this time, in 2011 recruitment?

Anybody on 23rd
 

mine is on 23rd anyone else having his GD on 23rd we can get in touch
[email protected]

GD Call letter
 

still waiting for the call letter

@aditya when did you clear your exam

Aug 2010,, I am from delhi

hope ppl started getting mails. Mine is on 2nd sept. Good luck guyss!!!

@ aviator_prat
 

How silly of me. I had got the Real drift. But for 45 degree north, I was calculating with hourly real drift and 45 mins of ERW.

my incorrect equation :

Md = RD + ERW
= 7.5 (increasing) + 15 * sin45 * (45/60)

This caused the blunder.

Anyways Thanks. Any idea for the remaining 1 question

Most awaited GD Date for Aug 2010 batch
 

Hey Fellow Aviators,
I just got Date for 2nd round of assessment 29th Aug.
All the very best ... :ok:

Finally got the e-mail . G.D on 26th . Good luck to everyone.:ok:

got 30th august for assessment center......I am from august 2010 batch......

All the best to all.....

got my date just now , i am from aug 2010 batch , hope u get yours soon

GD date
 

Got my GD on 29th August....all d best...

got mine on 30th as well.....all the best!!

guys who got it on the 29th n 30th are youll from delhi??