ATP Radio aids DGCA Prep Questions
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ATP Radio aids DGCA Prep Questions
Can anyone solve this and show how to get the answer.
An aircraft has to communicate with a VHF station at a range of 300 nm, if the ground station
is situated 2500' amsl which ofthe following is the lowest altitude at which contact is likely to
be made?
a) 190'
b) 1,378'
c) 37,600'
d) 84,100'
An aircraft at19,000ft wishes to communicate with a VDF station at 1,400ft amsl. What is the
maximum range at which contact is likely?
a) 175nm
b) 400.0nm
c) 62.5nm
d) 216nm
An aircraft has to communicate with a VHF station at a range of 300 nm, if the ground station
is situated 2500' amsl which ofthe following is the lowest altitude at which contact is likely to
be made?
a) 190'
b) 1,378'
c) 37,600'
d) 84,100'
An aircraft at19,000ft wishes to communicate with a VDF station at 1,400ft amsl. What is the
maximum range at which contact is likely?
a) 175nm
b) 400.0nm
c) 62.5nm
d) 216nm
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Question1
Range (nm) = 1.25 sq root ht. of transmitter + 1.25 sq. root ht of reciever
300 =1.25 sq root 2500 + 1.25 sq root ht of rec.
300 = 62.5 + 1.25 sq root ht of rec
300-62.5 = 1.25 sq root ht of rec
237.5 =1.25 sq root ht of rec
237.5/1.25= sq. root ht of rec ( Assume this to be X feet amsl )
190 = sq root x
sq. both sides
x= 36100 feet
Hence, the close option is C
Question2
Same formula to be used for this question, just that this tym range is to be calculated.
Range (nm) = 1.25 sq root 19000 + 1.25 sq. root 1400
= 172.30 +46.77
=219 nm approx
Hence, close option is D
Correct me if I'm wrong.!
Range (nm) = 1.25 sq root ht. of transmitter + 1.25 sq. root ht of reciever
300 =1.25 sq root 2500 + 1.25 sq root ht of rec.
300 = 62.5 + 1.25 sq root ht of rec
300-62.5 = 1.25 sq root ht of rec
237.5 =1.25 sq root ht of rec
237.5/1.25= sq. root ht of rec ( Assume this to be X feet amsl )
190 = sq root x
sq. both sides
x= 36100 feet
Hence, the close option is C
Question2
Same formula to be used for this question, just that this tym range is to be calculated.
Range (nm) = 1.25 sq root 19000 + 1.25 sq. root 1400
= 172.30 +46.77
=219 nm approx
Hence, close option is D
Correct me if I'm wrong.!
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You got the first one right but for some reason Oxford book is showing as second Question A as the answer. I do not know how they got to that.
I have a Quick Question if anyone knows this
If I hold an FAA ATP License, and I have to convert to DGCA ATP after holding Indian CPL and RT and FRTOL, I understand we have to give Composite ATP Nav/Radio Aids/Met and no Regs
Do we have to do viva also unlike General Issuance of ATP Paper where 3 seperate papers [Gen Navs, Radio Aids, Met and Viva] to get ATP
Please clarify
I have a Quick Question if anyone knows this
If I hold an FAA ATP License, and I have to convert to DGCA ATP after holding Indian CPL and RT and FRTOL, I understand we have to give Composite ATP Nav/Radio Aids/Met and no Regs
Do we have to do viva also unlike General Issuance of ATP Paper where 3 seperate papers [Gen Navs, Radio Aids, Met and Viva] to get ATP
Please clarify
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what is the rhumb line track from A (4500N 01000W) to B (4830N 01500W) ?
Thats all the info provided in the question
can't seem to get my head around it?
please post the complete answer with method.
Thanks!
Thats all the info provided in the question
can't seem to get my head around it?
please post the complete answer with method.
Thanks!
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okay i am not sure if my approach was right i just did what i could make out of this problem but i'l tell you what i did ....
try and make a triangle , with two corners as coordinates you have just mentioned and third this ( 45 N , 15 W ) .....so now its a basic triangle question ....find values of two sides one joining 45 N to 48 30 N and other side being calculated as 5 (ch long ) X 60 X cos mean lat .
Tan @ = Perpendicular / Base
you will get the angle as 45 something , add it 270 ( as 15 W is 270 of 10 W ) . It should add up to 315.2
SO GUYS WHAT DO YOU THINK IS IT THE RIGHT WAY OR I JUST GOT A FLUKE ???
try and make a triangle , with two corners as coordinates you have just mentioned and third this ( 45 N , 15 W ) .....so now its a basic triangle question ....find values of two sides one joining 45 N to 48 30 N and other side being calculated as 5 (ch long ) X 60 X cos mean lat .
Tan @ = Perpendicular / Base
you will get the angle as 45 something , add it 270 ( as 15 W is 270 of 10 W ) . It should add up to 315.2
SO GUYS WHAT DO YOU THINK IS IT THE RIGHT WAY OR I JUST GOT A FLUKE ???
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Thanks A lot bro!
I think its legitimate method, one of the other guys answered as well and he had the same method
post 523
http://www.pprune.org/south-asia-far...ml#post6404156
I think its legitimate method, one of the other guys answered as well and he had the same method
post 523
http://www.pprune.org/south-asia-far...ml#post6404156
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yup, thats good enough.
You can do Sin, Tan, Cos as long as you have the necessary variables like Opp, Base and Hypo. values in distances.
Good Job.
So draw the Isoceles Triangle with coordinates. Be careful where A and B points will lie and which way you are traveling in the right Rhumb Line Direction and then once found out that, then find out Coordinate C which is really simple
If A was 45 N 010 W, B is 48*30**, 15*W then C which is the side between Opp and Base of the triangle would be 45N 15W.
Then find distances of A to C using Dep= dlong X 60 X cos lat = 212NM
A-B (Hypo side) = 298.4NM and B-C = 210 NM
Then use Tan, or Sin from Point A. I took Sin
Sin A = 210/298.4 = 44.8 Degrees
Add 44.8+ 270 (A-C direction is going from 10*W to 15*W so you are going West) = 315
Yash you did good, I was just giving alternative for others if they may be confused or never done CPL exams.
You can do Sin, Tan, Cos as long as you have the necessary variables like Opp, Base and Hypo. values in distances.
Good Job.
So draw the Isoceles Triangle with coordinates. Be careful where A and B points will lie and which way you are traveling in the right Rhumb Line Direction and then once found out that, then find out Coordinate C which is really simple
If A was 45 N 010 W, B is 48*30**, 15*W then C which is the side between Opp and Base of the triangle would be 45N 15W.
Then find distances of A to C using Dep= dlong X 60 X cos lat = 212NM
A-B (Hypo side) = 298.4NM and B-C = 210 NM
Then use Tan, or Sin from Point A. I took Sin
Sin A = 210/298.4 = 44.8 Degrees
Add 44.8+ 270 (A-C direction is going from 10*W to 15*W so you are going West) = 315
Yash you did good, I was just giving alternative for others if they may be confused or never done CPL exams.
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@livetokill -
dude change your username , a pilot with a username live to kill is kind of scary
Jokes aside , as i saw your post in the Indigo forum , i'd just inform you similar to this quest was asked in indigo's exam .
dude change your username , a pilot with a username live to kill is kind of scary
Jokes aside , as i saw your post in the Indigo forum , i'd just inform you similar to this quest was asked in indigo's exam .
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Assuming the maximum likely error in VOR to be 5.50
, what is the maximum distance apart that
beacons can be situated on the centre line of an airway in order that an aircraft can guarantee
remaining within the airway boundary?
a) 54.5 nm
b) 109 nm
c) 66 nm
d) 132 nm
, what is the maximum distance apart that
beacons can be situated on the centre line of an airway in order that an aircraft can guarantee
remaining within the airway boundary?
a) 54.5 nm
b) 109 nm
c) 66 nm
d) 132 nm
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@ livetokill22
the airway width is 5 nm from center line .therefore at 5.5deg distance is = 5NM/Tan(5.50) = 51.92 NM the error is either + or - so 51.92x2= 103.85NM closest answer is 109NM.
The answer is 109nm, have seen the question in a book.
but don't know how they got 109nm
if someone could please elaborate.
Thanks!
but don't know how they got 109nm
if someone could please elaborate.
Thanks!
the airway width is 5 nm from center line .therefore at 5.5deg distance is = 5NM/Tan(5.50) = 51.92 NM the error is either + or - so 51.92x2= 103.85NM closest answer is 109NM.
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An aircraft is tracking away from a VOR on a heading of 287°M with 14° starboard drift. If the variation is 6°W what is the phase difference between the reference and variable phase components of the VOR transmission:
a) 121°
b) 295
c) 301 °
d) 315
a) 121°
b) 295
c) 301 °
d) 315
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An aircraft wishes to track towards a VOR along the 274 radial. If variation is 1O degree W what should be set on the OBS ?
a) 274
b) 264
c) 094
d) 084
a) 274
b) 264
c) 094
d) 084