ATP Radio aids DGCA Prep Questions
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It should be going from left to right and down a bit. Now put the rhumb line between A & B on the Equator side of the great circle you have drawn. We are told that the rhumb line course is 100?T. The sketch you have drawn indicates that the initial great circle track is less than 100?T and the final great circle track is more than 100?T. The average great circle track is 100?T and this will occur on the great circle at its mid-point, i.e. where it crosses 005?E.
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I initially got 104 too.
Is the trick of the question the fact that it says "The average true course of the great circle is 100°" ??
In that case 100 would be the GC track in between A to B.
Hence, GC at A would be 096 and GC on arrival at B would be 104. (Convergency being 8)
and since Conversion Angle is 4... RL at A would then be 100.
Is the trick of the question the fact that it says "The average true course of the great circle is 100°" ??
In that case 100 would be the GC track in between A to B.
Hence, GC at A would be 096 and GC on arrival at B would be 104. (Convergency being 8)
and since Conversion Angle is 4... RL at A would then be 100.
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Hey Yash , think of it like this..
>When you depart from point 'A' following the GC track towards 'B', the GC track is 096, and on reaching pont 'B' the GC track would be 104. (since Convergency is 8, and remember when flying eastward in the Northern Hemisphere GC track increases). So when you are at the midpoint on your GC journey from 'A' to 'B' , the GC track would be 100. (which is what they are trying to say when they mention the average GC track is 100)
>When you depart from point 'A' following the GC track towards 'B', the GC track is 096, and on reaching pont 'B' the GC track would be 104. (since Convergency is 8, and remember when flying eastward in the Northern Hemisphere GC track increases). So when you are at the midpoint on your GC journey from 'A' to 'B' , the GC track would be 100. (which is what they are trying to say when they mention the average GC track is 100)
Planeboy_777
In your statement
you appear to have assumed that the ratio of gradient / angle is a straight line. Having used 5 / 3 to get 1.66 for the 3 degree angle, you then suggest that this ratio applies for all angles. It does not.
For example for 2 degrees the gradient is 3.49% and the ratio of gradient / angle gives 3.49 / 2 = 1.746.
For 5 degrees the gradient is 8.75% and the ratio of gradient / angle gives 8.75 / 5 = 1.7498.
For 10 degrees the gradient is 17.63% and the ratio of gradeint / angle gives 1.763.
Your figure of 1.66 is in fact correct only for a 3 degree slope.
I do not know exactly what range of glide slopes for this type of question can be used in the DGCA examinations, but in the JAR exams they range for 3 degrees to 15 degrees.
Did you deduce the 1.66 figure for yourself or were you taught it by an instructor?
In your statement
1.666 x theta(deg) x Groundspeed(kts) = ROD(fpm) [[[[[For any GPA]]]]]
For example for 2 degrees the gradient is 3.49% and the ratio of gradient / angle gives 3.49 / 2 = 1.746.
For 5 degrees the gradient is 8.75% and the ratio of gradient / angle gives 8.75 / 5 = 1.7498.
For 10 degrees the gradient is 17.63% and the ratio of gradeint / angle gives 1.763.
Your figure of 1.66 is in fact correct only for a 3 degree slope.
I do not know exactly what range of glide slopes for this type of question can be used in the DGCA examinations, but in the JAR exams they range for 3 degrees to 15 degrees.
Did you deduce the 1.66 figure for yourself or were you taught it by an instructor?
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i have solved many problem with this formula and have successfully succeeded. Please dont question me...just substitute the values in correct units and get the answer....
my concept of gradient and angles are far more crystal clear than yours...
a radian expressed in percentage is a gradian
5.2% gradient is 5.2/100 radian
i.e 0.052 x 57.33 = 3 degrees
10% gradient means 10/100 radian
0.1 x 57.33 = 5.7 degrees
so according to the modified simple formula
1.666 x 5.7degrees x 100kts = 949.62
according to the formula given
5 x Grndsp x [5.7/3] = 950
dude go and revise your basic 6th grade maths....
my concept of gradient and angles are far more crystal clear than yours...
a radian expressed in percentage is a gradian
5.2% gradient is 5.2/100 radian
i.e 0.052 x 57.33 = 3 degrees
10% gradient means 10/100 radian
0.1 x 57.33 = 5.7 degrees
so according to the modified simple formula
1.666 x 5.7degrees x 100kts = 949.62
according to the formula given
5 x Grndsp x [5.7/3] = 950
dude go and revise your basic 6th grade maths....
Planeboy_777
As you grow older you will find that interaction with other people frequently results in them questioning you. They're not trying to be offensive, it's just the way that things are. Unless you can come to terms with this simple fact of life you are destined to have serious relationship problems in the future.
You have stated that
OK let's try it out for 45 degrees descent at 100 knots groundspeed.
If you sketch it out you will find that the ROD is equal to the groundspeed.
1 knot x 6080 ft/hour
Multiplying this by 1 hour / 60 minutes converts this into 101.3 ft/minute
multiplying by 100 knots gives ROD = groundspeed = 101.3 x 100 = 10130 ft/min.
So at 100 knots in a 45 degree glide our ROD = 10130 ft/min
Now let's try your formula
ROD = 1.66 x descent angle in degrees x groundspeed.
ROD = 1.66 x 45 x 100 knots groundspeed = 7470.
There is a considerable difference between 10130 ft/min and 7470 ft/min.
As I said in my earlier post, yout factor of 1.66 is Ok for 3 degrees, but becomes increasing less accurate as the angle increases.
Your definition of %gradient based on radians is Ok for maths exams, but in this forum we are tallking about aviation. A number of ATPL questions ask for the definition of %climb gradient. The correct answer is ROC/GS.
I have never yet seen an option involving radians in such questions.
Please dont question me...just substitute the values in correct units and get the answer....
You have stated that
1.666 x theta(deg) x Groundspeed(kts) = ROD(fpm) [[[[[For any GPA]]]]]
OK let's try it out for 45 degrees descent at 100 knots groundspeed.
If you sketch it out you will find that the ROD is equal to the groundspeed.
1 knot x 6080 ft/hour
Multiplying this by 1 hour / 60 minutes converts this into 101.3 ft/minute
multiplying by 100 knots gives ROD = groundspeed = 101.3 x 100 = 10130 ft/min.
So at 100 knots in a 45 degree glide our ROD = 10130 ft/min
Now let's try your formula
ROD = 1.66 x descent angle in degrees x groundspeed.
ROD = 1.66 x 45 x 100 knots groundspeed = 7470.
There is a considerable difference between 10130 ft/min and 7470 ft/min.
As I said in my earlier post, yout factor of 1.66 is Ok for 3 degrees, but becomes increasing less accurate as the angle increases.
Your definition of %gradient based on radians is Ok for maths exams, but in this forum we are tallking about aviation. A number of ATPL questions ask for the definition of %climb gradient. The correct answer is ROC/GS.
I have never yet seen an option involving radians in such questions.
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Alrite guys this formula has been derived from 1:60 rule. Note that the rule works reasonable well( to within 10%) for angles upto 20deg. If you tried to use the 1:60 rule with an angle of 45 deg, the estimation would not work.
Do not use the 1:60 rule for problems involving angles greater than 20deg.
Fortunately your glideslopes will be less than 20deg(for fixed wing a/c) and, hopefully, your track errors will not be greater than 20deg !
Given these conditions the formula can be applied:-
where he came up with 1.66 is
ROD= Glide Path Angle x 100 x G.S/60
ROD=1.667(100/60) x Glide path angle x G.S
For standard 3 deg glide path the formula is
ROD= G.S x 5
@KEITH WILLIMAS
Buddy I hope you knw the fact that this formula is applicable only till 20 deg so dont play with boy.
Do not use the 1:60 rule for problems involving angles greater than 20deg.
Fortunately your glideslopes will be less than 20deg(for fixed wing a/c) and, hopefully, your track errors will not be greater than 20deg !
Given these conditions the formula can be applied:-
where he came up with 1.66 is
ROD= Glide Path Angle x 100 x G.S/60
ROD=1.667(100/60) x Glide path angle x G.S
For standard 3 deg glide path the formula is
ROD= G.S x 5
@KEITH WILLIMAS
Buddy I hope you knw the fact that this formula is applicable only till 20 deg so dont play with boy.
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Guys if any one of you have appeared in the previous ATP exams can you please share what was the weightage given to topics like Magnetism , Gyros , INS and IRS .
Any help will be appreciated
Thanks
Any help will be appreciated
Thanks
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DME RIX (N56'53.8'' E23'58'')
RMI 164 degrees 48 NM DME
Position of aircraft is:
A) 57'48''N 023'43''E
B)57'42''N 023'40''E
C)57'42''N 023'50''E
D)57'50''N 023'38''E
RMI 164 degrees 48 NM DME
Position of aircraft is:
A) 57'48''N 023'43''E
B)57'42''N 023'40''E
C)57'42''N 023'50''E
D)57'50''N 023'38''E
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well the answer is simple, the error is 5.5 deg, and the width of an airway is 5 nm. hence apply the 1 in 60 rule and u get the distance from one end to the VOR as 55.54 nm. multiplying it by 2 you get 109 nm.
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ROD on a 3.5° Glideslope
Yash has suggested a correct solution.
Allow me to elaborate
1 Nm = 6080'
Ac at 120 Kts flies 2 Nm in a min = 12160'
Dist Off track (ROD)
---------------------- X 60 = Track error (Glide Path angle)
Dist to go (Dist to fly)
which is the same as
____________________ Dist to go (Dist to fly) X Track error (GP angle)
Dist Off track (ROD) = ----------------------------------------------
_______________________________________ 60
_________________ 12080' X 3.5 _____ 42280
Therefore (ROD) = --------------- = ---------- = 704.67'/min
___________________ 60_____________ 60
This might help those who are reading this forum for help if not the ones who posted much earlier.
Allow me to elaborate
1 Nm = 6080'
Ac at 120 Kts flies 2 Nm in a min = 12160'
Dist Off track (ROD)
---------------------- X 60 = Track error (Glide Path angle)
Dist to go (Dist to fly)
which is the same as
____________________ Dist to go (Dist to fly) X Track error (GP angle)
Dist Off track (ROD) = ----------------------------------------------
_______________________________________ 60
_________________ 12080' X 3.5 _____ 42280
Therefore (ROD) = --------------- = ---------- = 704.67'/min
___________________ 60_____________ 60
This might help those who are reading this forum for help if not the ones who posted much earlier.
Last edited by Ilyushin78MKI; 2nd Apr 2014 at 16:12.
