Indigo Call letters for Freshers

[AVIATROZ]

Q. you intend to overfly a mountain ridge at an altitude of 15000ft AMsL.The average air temp i s 15degree lover than IsA.the sealevel pressure 1003Hpa.which altimeter indication is needed?

Ans--16170ft...............please explain hw to solve this problem
thanks


please some one guide in solving this problem

[pilot in progress]

Hi everyone,

could anybody please help me with a couple of questions. Cant figure them out!!

1. The total length of the 53° N parallels of latitude on a direct Mercator chart is 133 cm. What is the approximate scale of the chart at latitude 30° S?

A) 1 : 25 000 000



2. On a Direct Mercator chart at latitude of 45° N, a certain length represents a distance of 90 NM on the earth. The same length on the chart will represent on the earth, at latitude 30° N, a distance of:

A) 110 NM.



Really appreciate for time and help..thankyou

[pilot in progress]

Q) apparent wander at 45N is:

[Flybi]

@alonelyaviator

how did u applied for the written?
only through cae website or by mailing the hr's also?

[sharma1prash]

hey
so thats how it works? first the candidate gets a mail from Radhika/Parul and you reply with the details.Till here i have the same story.Afterwards i have not got any kind of correspondence from CAE/Indigo for the assessment on 21st/22nd.so i guess that is not good?

[SaabCapt]

What is the UTC time of sunrise in Vancouver, British Columbia, Canada (49N 123 30 W) on the 6th December?
A)1552 UTC B)0724 UTC C)0738 UTC D)2324 UTC



Help! ..?

[mike_sierra]

1. The total length of the 53° N parallels of latitude on a direct Mercator chart is 133 cm. What is the approximate scale of the chart at latitude 30° S?

A) 1 : 25 000 000

For a Mercator, Scale at lat = Scale at equator * sec lat
in other words,
Denominator A/Denominator B = cos A/cos B = Earth dist A/Earth dist B

length of 53deg parallel is 360*60*cos 53=13000nm. Scale at 53 is 1:18112720
18112720/Denominator B = cos 53/cos 30
Denominator B =26,065,264
i.e. Scale at 30deg S is 1:26,065,264 (1:25,000,000 is the closest i suppose)

2. On a Direct Mercator chart at latitude of 45° N, a certain length represents a distance of 90 NM on the earth. The same length on the chart will represent on the earth, at latitude 30° N, a distance of:

A) 110 NM.

You could solve this using the same method.

[mike_sierra]

Q) apparent wander at 45N is:

Apparent Wander = 15deg * sin lat
at 45 N, apparent wander is 15*sin 45=10.6 deg/hr DECREASING

[pilot in progress]

@mike_sierra

check you pm

[9FLYJET9]

ANYONE WITH THE FINAL CALL LETTER FOR 21st or 22nd ?

[Percept]

No Hall Ticket or Venue Details yet, i guess the exam will be delayed or we will just get a surprise email on Monday

[9FLYJET9]

@percept
LOOKS LIKE.. LETS SEE..
WHEN DID YOU SEND A REPLY TO THE SECOND MAIL ?
DO POST ONCE YOU GET THE FINAL CALL LETTER..
TC..

[Percept]

@ 9FLYJET9

I received the second email on September 14th around noon, i sent a confirmation email to Capt Upendranath immediately.

I have heard that an exam previously was delayed by a couple of days - so that is a possibility.

But then again they can always send us a Hall ticket with Venue details on Monday, so lets keep studying!

A delay is good cuz then we have more time to prepare.

Best of luck to all of us.

[pilot in progress]

@percept

check your pm

[capt.neo]

Hey guys can u help me out

Q.)an a/c is intending the track from NDB A to NDB B on a track of 050(T),heading 060(T). If the RBI shows relative bearing of A to be 180 abd relative bearing of B to be 330 then the a/c is:

1.port of track and nearer A
2.port of track and nearer B
3.starboard of track and nearer A
4.starboard of track and nearer B

Ans:4

I figured out the starboard part of the track but can't understand how the a/c is nearer to B. Can someone help me out on this...???

[AVIATROZ]

thanks error detector

thanks bro don`t mind but i hav 1 more doubt for u


Q2.An OAT of -55degree at the 200Hpa pressure level:
A)+/-5degree IsA.
B) is low...............
{acc. to me ans should be +10degree IsA DEVIATION}
please advice regarding this problem

[errordetector]

@ AVIATROZ

True altitude = Indicated altitude + 4% of Indicated every 10 deg above ISA if ISA deviation is positive.
= Indicated altitude - 4% of Indicated every 10 deg below ISA if ISA deviation is negative.

Here 15000 is the true altitude. Temp is 15 deg lower so Indicated altitude will be higher. 4% for 10 deg so 6% for 15 deg.

6 % of 15000 is 900. So Indicated altitude is 15900.
The altimeter setting is 1003 so you will need to set 27 x10(1013-1003) = 270 feet higher.

15900 + 270 = 16170 feet which is your answer.

Hope you got it.

[Percept]

@ AVIATROZ

at 200mb pressure you are in an approximately at 38,600feet in the isothermal layer with a constant temperature of -56.5 degrees Celsius.

Since the OAT is -55, A)+/-5degree ISA should be the right answer.

[AVIATROZ]

hey thanks

ohhh .........so dumb of me not to think that silly think............thanks bro

a big thank to percept and error detector..........when no one replied to my problem u guys were life saver............keep it up guys keep helping each others

[SaabCapt]

On a 12% glide slope, your groundspeed is 540 knots. What is your rate of descent?
A)6550 feet/min. B)3120 feet/min. C)8740 feet/min. D)4820 feet/min.




Now how do we do this??