This is what I can make sense of. Unable to think beyond it.

Conversion angle=0.5*dlong *sin mean lat
=0.5*10*sin56
=4.145 degree(or 4 degrees)

so, initial GC track at waypoint 3(or 20 W)=(270+4)= 274 degrees

Now drift = angular difference between track and heading

HDG= 270 degrees (therfore drift= 4 degrees SB)

Similarly GC track at waypoint 4 (or 30 W) = (270-4)= 266 degrees

Here drift = 4 degrees port (assuming hdg remains 270)

i think the answer has to be "B"..
the heading is decreasing by approx 8.6 degrees..

cause when u depart from way point 3 your hdg is 274.3..
when you arrive at way point 4 your heading is 265.6 ...
so with constant drift throughout the flight...your hdd is decreasing by less than 10 degrees....i.e. 8.6...

please correct me if i am wrong...

Your method and answer are correct Wings of fire.

But your comment that

Is true when discussing drift in the normal sense, but this question isn't really about drift. I think that the examiner put in the words "constant drift" just to eliminate any unwanted complications. It would probably have been better to have said "in nil wind conditions".

The difference between the Great Circle track and the Rhumb Line track isn't realy an example of drift. At the start of this leg when the heading is 274.145, the track will also be 274.145 (assuming nil wind). And at the end of the leg when the heading was 265.855 the track will also be 265.855. But the average track over the whole leg would have been 270.

Thank you keith. I appreciate your sincere efforts.

ATPL on 19,20 and 21........all the best guys.....

guys! is there gonna be a list of admitted candidates any sooner?
or a VENUE to where to take the exam! Its already the 16th and 19th is the exam and no Venue and no admitted list and no nothing! OmG!

me too, waiting for the list and Venue! whats up with DGCA being so slow in declaring the listed candidates. Hardly any reaction time for those who dont make it in the list.

Well, Thats DGCA, Everything is fair in case of DGCA. The are the Brand-ambassadors of Laziness. :ugh:

@keith williams
 

Hello keith. Can u solve this question. Again from radio aids oxford book- vor chapter.


Q4)what is the theoretical maximum range that an aircraft at FL 420 will obtain from a vor beacon situated at 400 feet above msl?
a) 225 nm
b) 256 nm
c) 281 nm
d) 257 nm

answer is a) 225 NM

My explanation:

Range = 1.25*(Root of ht + root of hr)

=1.25*(root of 400 + root of 42000)

= 281 NM

Your soluton is correct.

The author appears to have forgotten to multiply by 1.25.

This kind of problem would be avoided if the book included worked solutions for each question, but this obviously involves more work for the author.

do all of you who applied,have your name in the admitted candidates list?

i dont understand, my form , draft were as per guidelines but still i didnt get my roll no.:ugh::ugh::ugh::ugh:

and they are saying

Application of those candidates whose name does not appear in the
above list has been rejected either due to incomplete application/not enclosing
the required documents/wrong demand draft. CEO shall entertain no
communication from the candidates, whose applications have been rejected.

What to do now????:confused::confused:

So the three venues of Delhi are!

JASOLA, DWARKA & ROHINI! Looks like DGCA is on a Pilots DILLI DARSHAN Mission!

No idea why such absurd locations were chosen! All three are in Different corners of the City!

One Centralized Location with different Timings for Exams over 3-4 days could have been done! Would have been easier for people not familiar with the City coming from outside!

On top of that they post the list of names just 2 days before the exam!
Don't want to entertain any issues of names they haven't put in the list!

Dictatorship General of Civil Aviation, India! :ugh: :ouch:

Next punch would be watching the systems crash!

same here buddy...the clowns havent even posted the rejection list so one doesnt even know what went wrong :mad:

@ Roll No.
 

Does anybody have a idea that do we have to go for stamping from DGCA official as in CPL exam or just a print out of roll no. will work..

Please Revert

The great circle track X - Y measured at x is 319°, and Y 325° Consider the following statements:
a Southern hemisphere, Rhumb line track is 322°
b Northern hemisphere, Rhumb line track is 313°
c Southern hemisphere, Rhumb line track is 331°
d Northern hemisphere, Rhumb line track is 322°


A great circle track joins position A (59°S 141°W) and B (61°S 148°W). What is the difference between the great circle track at A and B?
a It increases by 6°
b It decreases by 6°
c It increases by 3°
d It decreases by 3°


Thanks

a Southern hemisphere, Rhumb line track is 322°

b It decreases by 6°

Damn i still remember :D:D

All the best guys for your ATPL:ok::ok:

The first ones right.

But the second ans. is given as Increase by 6

When taking the exam the most important thing is to select the correct answer.

When preparing to take the exam the most important thing is to learn the correct method.


The important thing to remember here is because both the rhumb line and the great circle join the two points, the rhumb line track is the average of the great circle track at the two points. So in this question we just need to add the two together and divide by two to get (319 + 325) / 2 = 322.

Great circles are always convex to the nearest pole.

We have a great circle that is changing from 319 to 325, so we must be in the southern hemisphere. Sketch the situation if have difficulty visualising it.


For this one we need to use the conversion angle formula

The conversion angle is the difference between the great circle and the rhumb line.

Conversion angle = 1/2 x Sin mean latitude x difference in longitude.

In this case we have

Mean latitude = (59S + 61S) / 2 = 60S.

So Conversion angle = 1/2 x Sin60S x ( 148W - 141W)

If we use the conventions that

Latitudes North are positive
Latitudes South are negative
Differences in longitude to the West are positive
Differences in longitude East are negative

We get

Conversion angle = 1/2 x Sin -60 x 7W = -3.03 degrees.

The conversion angle is the difference between the great circle and the rhumb line so at position A the great heading is 3.03 degrees less than the rhumb line.

But the great circle and the rhumb line meet at positioon B so the great circle heading must gradually increase until it becomes 3.03 degrees greater than the rhumb line.

Starting 3.03 less and ending up 3.03 more, the great circle must have increased by 6.06 degrees.

If the use of plus and minus values is too confusing, then simply sketch the situation.

Great circles are always convex to the nearest pole.

We are in the southern hemisphere so the great circle initialy heads south of the starting point then gradually curves to the north until it reaches the second position. We are going west so this curving must represent a gradual increase in heading.

So the answer to this question is increasing by 6.06 degrees.

Thank you sir.

Truly genius. :D:D:D

Please also could you clarify.

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 45 NM
b 73.5 NM
c 78 NM
d 110 NM

&


The standard parallels of a Lambert's conical orthomorphic projection are 07°40'N and 38°20' N. The constant of the cone for this chart is:
a 0.60
b 0.39
c 0.92
d 0.42

@azax
Dude seriously .. I would suggest put some effort..and you will get it...

Good luck to all appearing for ATPL

@ i did get the 1st one later..
cos30/cos45 * 90 nm = 110.2 nm

but i have no clue of 'constant of cone'
if you do know..you might as well post the solution