JAR ATPL Feedback

[Midland Maniac]

This is an appeal for anyone who sat the ATPL's last week!!!

I am trying to get down all of the feedback questions that I can remember from the following exams: -

General Nav
Radio Nav
Opps Proc
Met
Instruments
Flight Planning

If anyone has noted some of the questions down, could you please send them on too me and in return, I will email you with the finished article. I need this info as I have a funny feeling that I may be doing a couple of re-sits!!!!!!! :eek: :eek: :eek:

[FatFlyer]

One nav question I had no idea about and would appreciate help for the resit-

Aircraft using INS to go from 60S 010W to 60S 020W what is the latitude at 015W
a) 60S
b) 59'50S
c) 60'06S
d 60'11S

using INS it is a great circle track so a and b are wrong , help

[Megaton]

How about this:

Distance from pole to 60 S is (30*60)=1800nm.
Distance from pole to track at mid=point is 1800 cos 5 since change in longitude=10 therefore ch long=5. Therefore distance from 60 S to midpoint of track is 1800-1800cos5=6 nm. Therefore lat = 60 deg 6 min.

Or is this a load of bollox?

[Tee]

MM

Some bits and pieces from August Operational Procedures exam sent by private e-mail

Tee

[thecoltster]

eaaasy?!!!

[sushi]

Hey everybody!
I'm doing the ATPL course and I have my exams in 2 months !!! I'm looking for JAA questions and would be happy if you have some info regarding questionaire data bases / collections.

Thx in advance.
Sushi

[BEagle]

c is the correct answer to the question above.

[Megaton]

That's what I said!

[Spartacus73]

Fat Flyer,

First posting!

Find conversion angle:
1/2 Ch.Long x Sin Lattitude

Work out departure and divide by 2:
Dep = Ch.Long(min)x Cos Lat

use 1 in 60 or trig to get disatnce
should be 5.4nm from a C.A. of 2.16 & distance of 150nm.

This is close enough apparently - spheroidal trigonometry not required - as this is not even in the learning objectives. Thats what my ground school instructor said.

Good luck!! :)

[Send Clowns]

Sparticus is almost exactly correct. Just use half the conversion angle, not all.

This is because at the point you are looking at the great circle (track) and rhumb line (parallel of latitude) are now parallel. If you want more detail email me (see my profile) and I will sketch a diagram and return it, as this is impossible to explain without a diagram!

So conversion angle is
1/2 ch.long. sin lat = 1/2 x 10 x sin 60
= 4.33
1/2 conversion angle = 2.17

Departure to mid-long = 60 x ch. long x cos lat
= 60 x 5 x cos 60
= 150 nm

1 in 60 rule : 150 is 2.5 x 60, so distance is 2.5 x angle
2.5 x 2.17 = 5.43 nm

so I would say the 60 degrees 6' North is best guess.

[ 12 August 2001: Message edited by: Send Clowns ]

[Midland Maniac]

Well that is just great isn't it !!! I ask for feedback and people tell me about qustions that I got wrong!!!! I couldn't do that one either....!!! May need to learn how for the resit!!!

Send Clowns!!! are you sure that you taught me how to do that one, because i couldnt remember in the exam!!!

MORE FEEDBACK QUESTIONS PLEASE!!!!! :rolleyes: :rolleyes: :rolleyes:

[Send Clowns]

No, SW taught you that one. I know because one of you lot used the technique in front of me, and said so!

[Scratch One Bandit]

Who? Me? :D

[Midland Maniac]

Well if he did, i can't remember!! I guess that I shall have to get him to show me that again when I eventually return to SFT!!!!

How is life down in Bournemouth anyway??? :D

[Send Clowns]

Not bad - very busy. Got the CAA in for inspection, and trying to sort out something to help PPSC's students. How did the exams go?

Are you SW, Scratch One??? Very sad location in your profile, though possibly quite correct.

[Scratch One Bandit]

Sorry, maybe that last post was a bit confusing. I'm the student who spoke to you about the INS latitute question above. And gave you the feedback on Fri. Yeah, I'm him.

Midland Maniac, I've wroten a little bit of feedback for GS, not much, but I guess he will have spoken to others since the exams.

And yeah, ok, I've changed my FROM info back to Bournemouth. :p