india DGCA ATPL exam
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@sbflyingangel: An attested copy of the CPL is the only document required for the ATPL application. Also as per the DGCA notice your documents need to reach the DGCA office within 10 days of you applying online. The last date to apply online was today (23rd Apr). Hope that helps!
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Thanks for the info
Thanku all for the quick reply...appreciate it.
I was able to figure it out on my own though..it just took me a while to get thru DGCA website..
@bayblade
i had already applied online...23rd was the last date to apply online.I had doubt regarding the docs to be sent..that is clear now.
I was able to figure it out on my own though..it just took me a while to get thru DGCA website..
@bayblade
i had already applied online...23rd was the last date to apply online.I had doubt regarding the docs to be sent..that is clear now.
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this is one page dedicated to me!
yes i'm sorry about the info, that is how i perceived it and i relayed it. it was incorrect, my bad
i was just trying to help.
good luck to all who apply!
yes i'm sorry about the info, that is how i perceived it and i relayed it. it was incorrect, my bad
i was just trying to help. good luck to all who apply!
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some problems...
1) An island is observed to be 15° to the left.The aircraft heading is 120°(M), variation 17°(W).The bearing °(T) from the aircraft to the island is:
A 302 B 088 C 122 D 268
2) A Lambert conformal conic chart has a constant of the cone of 0.75.The initial course of a straight line track drawn on this chart from A (40°N 050°W) to B is 043°(T) at A; course at B is 055°(T).What is the longitude of B?
A 36°W B 38°W C 41°W D 34°W
3) On a chart, the distance along a meridian between latitudes 45°N and 46°N is 6 cm. The scale of the chart is approximately:
A 1 : 1 850 000 B 1 : 1 000 000 C 1 : 185 000 D 1 : 18 500 000
4) A ground feature appears 30° to the left of the centre line of the CRT of an airborne weather radar. If the heading of the aircraft is 355° (M) and the magnetic variation is 15° East, the true bearing of the aircraft from the Feature is:
A 310° B 130° C 160° D 220°
5) The constant of cone of a Lambert conformal conic chart is quoted as 0.3955.At what latitude on the chart is earth convergency
correctly represented?
A 66°42' B 68°25' C 21°35' D 23°18'
A 302 B 088 C 122 D 268
2) A Lambert conformal conic chart has a constant of the cone of 0.75.The initial course of a straight line track drawn on this chart from A (40°N 050°W) to B is 043°(T) at A; course at B is 055°(T).What is the longitude of B?
A 36°W B 38°W C 41°W D 34°W
3) On a chart, the distance along a meridian between latitudes 45°N and 46°N is 6 cm. The scale of the chart is approximately:
A 1 : 1 850 000 B 1 : 1 000 000 C 1 : 185 000 D 1 : 18 500 000
4) A ground feature appears 30° to the left of the centre line of the CRT of an airborne weather radar. If the heading of the aircraft is 355° (M) and the magnetic variation is 15° East, the true bearing of the aircraft from the Feature is:
A 310° B 130° C 160° D 220°
5) The constant of cone of a Lambert conformal conic chart is quoted as 0.3955.At what latitude on the chart is earth convergency
correctly represented?
A 66°42' B 68°25' C 21°35' D 23°18'
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1) Hdg = 120 (M)
Island is 15 degrees to left.
Hdg to Island = 105 deg
T V M D C
88 17W 105
Thus, True brg to Island is 88 deg
2) conv = 12 deHdg tg (55-43)
conv = dlong x sin lat
since constant of cone is 0.75 , sin lat = 0.75
Thus, dlong = 12/0.75 = 16 deg
( since course is 043 which means the a/c is moving east)
50 W - 16 = 34 W
3) Scale formula - 6 cm /[ 1 (deg) x 60 (nm) x 185200 cm]
= 1/1 852 000 ~ 1/1 850 000
{1 deg = 60 nm; 1 nm = 1.852 km = 1852 m = 185200 cm}
4) Same as 1) Hdg to ground feature = 355 - 20 = 325 M
Apply CDMVT, True hdg to ground feature = 340 deg
Hdg OF a/c from ground feature = 340 - 180(reciprocal)
= 160 deg
5) contant of cone = sin (lat)
sin (lat) = 0.3955
Thus, lat = 23 deg 18 minutes
Island is 15 degrees to left.
Hdg to Island = 105 deg
T V M D C
88 17W 105
Thus, True brg to Island is 88 deg
2) conv = 12 deHdg tg (55-43)
conv = dlong x sin lat
since constant of cone is 0.75 , sin lat = 0.75
Thus, dlong = 12/0.75 = 16 deg
( since course is 043 which means the a/c is moving east)
50 W - 16 = 34 W
3) Scale formula - 6 cm /[ 1 (deg) x 60 (nm) x 185200 cm]
= 1/1 852 000 ~ 1/1 850 000
{1 deg = 60 nm; 1 nm = 1.852 km = 1852 m = 185200 cm}
4) Same as 1) Hdg to ground feature = 355 - 20 = 325 M
Apply CDMVT, True hdg to ground feature = 340 deg
Hdg OF a/c from ground feature = 340 - 180(reciprocal)
= 160 deg
5) contant of cone = sin (lat)
sin (lat) = 0.3955
Thus, lat = 23 deg 18 minutes
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@captaa
Are you serious ? Maybe you did not understand what keith was trying to say. Everyone pays for their respective internet charges buddy, no big deal there. When you download something which is not available for free (here i am referring to the charges of the product and not the internet charges), you are actually stealing someone's work.
To make it simpler for you, suppose you dedicate a major time of your life writing a book, which was never intended to be sold for free of course. Then comes some
person who purchase the copy of your book, scans it and upload it on the internet. Now the whole world downloads your book from the internet FOR FREE and then when you look at the statistics, you're shocked to see the whole world using your book but total sales of your book is very less comparitively. That's stealing someone's hard work, also known as piracy.
Too difficult to understand ???
Are you serious ? Maybe you did not understand what keith was trying to say. Everyone pays for their respective internet charges buddy, no big deal there. When you download something which is not available for free (here i am referring to the charges of the product and not the internet charges), you are actually stealing someone's work.
To make it simpler for you, suppose you dedicate a major time of your life writing a book, which was never intended to be sold for free of course. Then comes some
person who purchase the copy of your book, scans it and upload it on the internet. Now the whole world downloads your book from the internet FOR FREE and then when you look at the statistics, you're shocked to see the whole world using your book but total sales of your book is very less comparitively. That's stealing someone's hard work, also known as piracy.Too difficult to understand ???
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Guys whats the best way to study for DGCA ATPL exams
I am giving Radio aids & Instruments, Meteorlogy...
The material I have = Bristol CBT and Keith williams 1000 questions
I am giving Radio aids & Instruments, Meteorlogy...
The material I have = Bristol CBT and Keith williams 1000 questions
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plz help....
1) r/l track btw a(45n 10w) and b(48'30n 15w) is
a)345 b) 315 c) 330 d) 300
2) a(55n 00) b (54n 10e). Avg true course of g/c is 100 deg. True crs of r/l at point a is
a) 100 b) 096 c) 104 d) 107
3) an a/c travels a g/c track from 56n 70w to 62n 110e. Total dist travelled is
a) 5420nm b) 1788nm c)2040nm d)3720nm
4)psn a is located on equtor at long. 130e psn b is located 100nm from a on a bearing of 225 deg (t). The co-ordinates of b are
a) 01'11's 131'11'e b) 01'11'n 128'49'e c) 01'11's 128'49'e d) 01'11'n 131'11'e
5) g/c dist. Btw psn a (59'34.1'n 008'08.4'e) and b (30'25.9'n 171'051.6'w)
is
a) 10800nm b) 5400nm c) 10800km d) 2700nm
a)345 b) 315 c) 330 d) 300
2) a(55n 00) b (54n 10e). Avg true course of g/c is 100 deg. True crs of r/l at point a is
a) 100 b) 096 c) 104 d) 107
3) an a/c travels a g/c track from 56n 70w to 62n 110e. Total dist travelled is
a) 5420nm b) 1788nm c)2040nm d)3720nm
4)psn a is located on equtor at long. 130e psn b is located 100nm from a on a bearing of 225 deg (t). The co-ordinates of b are
a) 01'11's 131'11'e b) 01'11'n 128'49'e c) 01'11's 128'49'e d) 01'11'n 131'11'e
5) g/c dist. Btw psn a (59'34.1'n 008'08.4'e) and b (30'25.9'n 171'051.6'w)
is
a) 10800nm b) 5400nm c) 10800km d) 2700nm
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1) r/l track btw a(45n 10w) and b(48'30n 15w) is
RL track from A to B is 270 + x deg
In triangle AOB
departure dist AO = d long in min x cos 45 = 5 x 60 x cos 45 = 212 nm.
great circle dist BO = (48.5 - 45) x 60 = 210 nm
tan x = BO/AO = 210/212
x = tan inv 210/212 = 44.7 deg
RL track = 270 + 44.7 = 314.7 deg
RL track from A to B is 270 + x deg
In triangle AOB
departure dist AO = d long in min x cos 45 = 5 x 60 x cos 45 = 212 nm.
great circle dist BO = (48.5 - 45) x 60 = 210 nm
tan x = BO/AO = 210/212
x = tan inv 210/212 = 44.7 deg
RL track = 270 + 44.7 = 314.7 deg
Last edited by gAMbl3; 29th Apr 2012 at 09:50. Reason: Typo
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3) an a/c travels a g/c track from 56n 70w to 62n 110e. Total dist travelled is
70 W and 110 E are meridian - anti meridian
A/c will fly from 56 N to 62 N via North pole
Gc dist from 56 N to north pole = (90 - 56) x 60 = 2040 nm
Gc dist from north pole to 62 N = (90 - 62) x 60 = 1680 nm
Total dist = 2040 + 1680 = 3720 nm
70 W and 110 E are meridian - anti meridian
A/c will fly from 56 N to 62 N via North pole
Gc dist from 56 N to north pole = (90 - 56) x 60 = 2040 nm
Gc dist from north pole to 62 N = (90 - 62) x 60 = 1680 nm
Total dist = 2040 + 1680 = 3720 nm
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4)psn a is located on equtor at long. 130e psn b is located 100nm from a on a bearing of 225 deg (t). The co-ordinates of b are
x = 225 - 180 = 45 deg
In triangle AOB
x = 45 deg
AO = OB (45 45 90 triangle)
dist AB = 100 nm
dist AO = dist OB AB Cos x = 100 x cos 45 = 70.7 nm
GC dist AO = 70.7 nm = d lat x 60
d lat = 70.7/60 = 1 deg 11 min
Latitude of B = 1 deg 11 min South.
departure dist OB = 70.7 nm = d long x 60 x cos lat
70.7 = d long x 60 x cos 1 deg 11 min = d long x 59.98
d long = 70.7/59.98 = 1 deg 10 min
Longitude of B = 130 E - 1 deg 10 min = 128 deg 49 min East.
Co ordinates of B (1*11' S 128*49' E)
x = 225 - 180 = 45 deg
In triangle AOB
x = 45 deg
AO = OB (45 45 90 triangle)
dist AB = 100 nm
dist AO = dist OB AB Cos x = 100 x cos 45 = 70.7 nm
GC dist AO = 70.7 nm = d lat x 60
d lat = 70.7/60 = 1 deg 11 min
Latitude of B = 1 deg 11 min South.
departure dist OB = 70.7 nm = d long x 60 x cos lat
70.7 = d long x 60 x cos 1 deg 11 min = d long x 59.98
d long = 70.7/59.98 = 1 deg 10 min
Longitude of B = 130 E - 1 deg 10 min = 128 deg 49 min East.
Co ordinates of B (1*11' S 128*49' E)