JAR ATPL Feedback
Guest
Posts: n/a
JAR ATPL Feedback
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!!!!!!!
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!!!!!!!
Join Date: Sep 2000
Location: London, England
Posts: 185
Likes: 0
Received 0 Likes
on
0 Posts
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
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
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?
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?
Join Date: Aug 2001
Location: weil am rhein
Posts: 1
Likes: 0
Received 0 Likes
on
0 Posts
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
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
That's what I said!
Join Date: Aug 2001
Location: Earth
Posts: 3
Likes: 0
Received 0 Likes
on
0 Posts
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!!
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!!
Jet Blast Rat
Join Date: Jan 2001
Location: Sarfend-on-Sea
Age: 51
Posts: 2,081
Likes: 0
Received 0 Likes
on
0 Posts
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 ]
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 ]
Guest
Posts: n/a
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!!!!!
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!!!!!
Jet Blast Rat
Join Date: Jan 2001
Location: Sarfend-on-Sea
Age: 51
Posts: 2,081
Likes: 0
Received 0 Likes
on
0 Posts
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.
Are you SW, Scratch One??? Very sad location in your profile, though possibly quite correct.
Join Date: Sep 2000
Location: Bournemouth, UK
Posts: 38
Likes: 0
Received 0 Likes
on
0 Posts
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.
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.
