Jet Airways Trainee First Officer
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@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?
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?
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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.)
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.
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Hence TAS will be taken care of due to additional increase of density due to decreasing temp.
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@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)
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
a) 92.33 degree
b)102.50 degree
c)97.78 degree
d) 103.75 degree
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)
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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?
And does anyone know how many vacancies Jet is looking to fill this time, in 2011 recruitment?
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Anybody on 23rd
mine is on 23rd anyone else having his GD on 23rd we can get in touch
[email protected]
[email protected]
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@ 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
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
