Indigo Call letters for Freshers
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@KKARR
A/c is corrected for apparent wander at 30 deg S and a/c is flying at 60 deg S
Lat nut correction = 7.5 decreasing (Since apparent wander at 30S is 7.5 inc.)
Apparent wander at 60 S = 12.99 increasing
Apparent wander experienced by a/c = 12.99 inc – 7.5 dec
= 5.49 inc
Total wander mentioned above = 21 inc.
Total wander = Transport wander + Apparent wander
Transport wander = 21 – 5.49 = 15.51
Transport wander = departure X tan lat.
Dept = (15.51/ tan 60) X 60
G/S = 537.2 KTS
Lat nut correction = 7.5 decreasing (Since apparent wander at 30S is 7.5 inc.)
Apparent wander at 60 S = 12.99 increasing
Apparent wander experienced by a/c = 12.99 inc – 7.5 dec
= 5.49 inc
Total wander mentioned above = 21 inc.
Total wander = Transport wander + Apparent wander
Transport wander = 21 – 5.49 = 15.51
Transport wander = departure X tan lat.
Dept = (15.51/ tan 60) X 60
G/S = 537.2 KTS
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@kickass aviator
cheers. appreciate it. also would any of you guys know what type of performance should be studied, would the performance in ace the techincal pilot interview be enough?
cheers. appreciate it. also would any of you guys know what type of performance should be studied, would the performance in ace the techincal pilot interview be enough?
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P.1.An aircraft heading 130° (M) has an ADF reading of 190° Relative. The heading to steer to intercept the 170° track outbound from the NDB at 30º is: A)190° (M) B)220° (M) C)200° (M) D)210° (M)
P.2.
An aircraft heading 040 (M) has an ADF reading of 060 Relative. The heading to steer to intercept the 120° track inbound to the NDB at 50° is:
A)060(M) B)070(M) C)080(M) D)050(M)
P.3.A hyperbola cuts the base line 60 Km from the Master end and 150 Km from the Slave end. When on the same hyperbola at a range of 90 Km from the Master, the range from the Slave will be:
A)300 km B)150 km C)240 km D)180 km
Anyone ?
P.2.
An aircraft heading 040 (M) has an ADF reading of 060 Relative. The heading to steer to intercept the 120° track inbound to the NDB at 50° is:
A)060(M) B)070(M) C)080(M) D)050(M)
P.3.A hyperbola cuts the base line 60 Km from the Master end and 150 Km from the Slave end. When on the same hyperbola at a range of 90 Km from the Master, the range from the Slave will be:
A)300 km B)150 km C)240 km D)180 km
Anyone ?
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P.1.An aircraft heading 130° (M) has an ADF reading of 190° Relative. The heading to steer to intercept the 170° track outbound from the NDB at 30º is:
Heading = 130 M
Rel Bearing = 190
QDM = Heading + Rel Bearing = 130 + 190 = 320
QDR = reciprocal of QDM = 320 -180 = 140. i.e a/c is on the 140 radial/outbound track and heading 130.
If a/c turns 10 degree right it's heading and outbound track will be equal(140 M), if it again turns 30 degree right(170 M) it will fly parallel to 170 outbound track and another 30 degree right (as it has to intercept at 30) it will intercept the 170 track at 30 degree. So it has to turn a total of 10+30+30 =70 degrees to right
Hence heading to steer = 130 M + 70 = 200 M
Similarly P2
Heading = 130 M
Rel Bearing = 190
QDM = Heading + Rel Bearing = 130 + 190 = 320
QDR = reciprocal of QDM = 320 -180 = 140. i.e a/c is on the 140 radial/outbound track and heading 130.
If a/c turns 10 degree right it's heading and outbound track will be equal(140 M), if it again turns 30 degree right(170 M) it will fly parallel to 170 outbound track and another 30 degree right (as it has to intercept at 30) it will intercept the 170 track at 30 degree. So it has to turn a total of 10+30+30 =70 degrees to right
Hence heading to steer = 130 M + 70 = 200 M
Similarly P2
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can anybody solve these?
A particular SID requires a minimum climb gradient of 210 feet per NM to 8,000 feet. If you climb with a groundspeed of 140 knots, what is the rate of climb required in feet per minute?
A) 304
B) 210
C) 450
D) 490
A sailplane has lost 2.000 feet in 9 nautical miles. The best glide ratio for this sailplane is approximately:
A) 24:1
B) 27:1
C) 30:1
D) 18:1
A particular SID requires a minimum climb gradient of 210 feet per NM to 8,000 feet. If you climb with a groundspeed of 140 knots, what is the rate of climb required in feet per minute?
A) 304
B) 210
C) 450
D) 490
A sailplane has lost 2.000 feet in 9 nautical miles. The best glide ratio for this sailplane is approximately:
A) 24:1
B) 27:1
C) 30:1
D) 18:1
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1) The rate of climb can be calculated manually. Take your ground speed in nautical miles per hour, divide by 60 minutes per hour, and multiply by the climb gradient in feet per nautical mile. The result will be the required rate of climb in feet per minute.
So, the ROC reqd. for the above question will be:
140 knots divided by 60= 2.33333 nautical miles per minute
2.33333 x 210 ft/NM = 490 feet/ min
2) Glide Range= (L/D) x height
Substituting the values we get (L/D)= 27:1
So, the ROC reqd. for the above question will be:
140 knots divided by 60= 2.33333 nautical miles per minute
2.33333 x 210 ft/NM = 490 feet/ min
2) Glide Range= (L/D) x height
Substituting the values we get (L/D)= 27:1
Join Date: Apr 2012
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Sad yaar, I have been mailing them from ages. first time they sent me initial mail asking for details. Never got roll number. Parul or Sandra never reply to mails nor pick up the calls...
Guys can u plz tell what can be done about this.
Guys can u plz tell what can be done about this.
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Anybody up for this, any help would be appreciated.
At 1000 Z an aircraft is overhead NB PE enroute to NDB CN, Track 075(M), Heading 082(M) At 1029 Z NDB PE bears 176 Relative and NDB CN bears 353 Relative. The heading to steer at 1029 Z to reach NDB CN is:
A) 078(M)
B) 081(M)
C) 0082(M)
D) 079(M)
At 1000 Z an aircraft is overhead NB PE enroute to NDB CN, Track 075(M), Heading 082(M) At 1029 Z NDB PE bears 176 Relative and NDB CN bears 353 Relative. The heading to steer at 1029 Z to reach NDB CN is:
A) 078(M)
B) 081(M)
C) 0082(M)
D) 079(M)
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An aircraft heading 135° (M) with 13° Right drift intercepts the 082° (M) track outbound from an NDB. The relative bearing of the NDB that confirms track interception is:
A) 137° Relative.
B) 132° Relative.
C) 122° Relative.
D) 127° Relative.
A) 137° Relative.
B) 132° Relative.
C) 122° Relative.
D) 127° Relative.
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At 1000 Z an aircraft is overhead NB PE enroute to NDB CN, Track 075(M), Heading 082(M) At 1029 Z NDB PE bears 176 Relative and NDB CN bears 353 Relative. The heading to steer at 1029 Z to reach NDB CN is:
Track 075 M
Heading 082 M
Wind correction angle = 82-75 = 7 deg to the right.
At 1029 Z
NDB PE
Heading + Relative bearing = QDM +/- 180 = QDR
082 + 176 = 258 -180 = 078 QDR
Plan track = 075
Track made good = 078
A/c has drifted 3 deg to the right (78 -75) which means the wind correction angle of 7 deg is more than that required for track 075 M. WCA should be 7 -3 = 4 deg.
NDB PE
Heading + Relative bearing = QDM
082 + 353 = 075 M QDM
To reach NDB PE the A/c needs to track 075M. Wind correction angle for 075 M should be 4 deg as explained above and so the heading to steer should be 075 + 4 = 079M
Track 075 M
Heading 082 M
Wind correction angle = 82-75 = 7 deg to the right.
At 1029 Z
NDB PE
Heading + Relative bearing = QDM +/- 180 = QDR
082 + 176 = 258 -180 = 078 QDR
Plan track = 075
Track made good = 078
A/c has drifted 3 deg to the right (78 -75) which means the wind correction angle of 7 deg is more than that required for track 075 M. WCA should be 7 -3 = 4 deg.
NDB PE
Heading + Relative bearing = QDM
082 + 353 = 075 M QDM
To reach NDB PE the A/c needs to track 075M. Wind correction angle for 075 M should be 4 deg as explained above and so the heading to steer should be 075 + 4 = 079M
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An aircraft heading 135° (M) with 13° Right drift intercepts the 082° (M) track outbound from an NDB. The relative bearing of the NDB that confirms track interception is:
Heading = 135 M
At the point of interception of 082 M outbound track :
If A/c heading = 082 M then Rel bearing = 180
But heading = 135 M
So rel bearing = 180 - (135-82) = 180 - 53 = 127.
Heading = 135 M
At the point of interception of 082 M outbound track :
If A/c heading = 082 M then Rel bearing = 180
But heading = 135 M
So rel bearing = 180 - (135-82) = 180 - 53 = 127.
Join Date: Jan 2012
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The great circle track X - Y measured at x is 319° , and Y 325° Consider the following statements
A) Northern hemisphere, Rhumb line track is 322° .
B) Northern hemisphere, Rhumb line track is 313° .
C) Southern hemisphere, Rhumb line track is 331° .
D) Southern hemisphere, Rhumb line track is 322° .
Guys any idea how to solve this one...???
A) Northern hemisphere, Rhumb line track is 322° .
B) Northern hemisphere, Rhumb line track is 313° .
C) Southern hemisphere, Rhumb line track is 331° .
D) Southern hemisphere, Rhumb line track is 322° .
Guys any idea how to solve this one...???
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indigo!!
Hello everybody, people who have got confirmation mails - Do you have current IR?
This is from CAE website.
"Indian Nationality Indian DGCA issued CPL or higher license with Multi Engine Instrument Rating Indian COP/RTR and FRTO Licenses must be valid at all times Valid Indian Class I medical certificate Fluency in English"
Thank You
Ayush
This is from CAE website.
"Indian Nationality Indian DGCA issued CPL or higher license with Multi Engine Instrument Rating Indian COP/RTR and FRTO Licenses must be valid at all times Valid Indian Class I medical certificate Fluency in English"
Thank You
Ayush